[proofplan] We first prove two elementary incidence facts: every line has an external point, and every two distinct lines have a point outside their union. The key counting comparison is that, for a point $p$ not on a line $\ell$, the lines through $p$ are in bijection with the points of $\ell$. This makes all line sizes equal, then all point degrees equal by the same comparison. Finally, fixing one point and counting the lines through it gives $|P|$, while the dual count gives $|L|$. [/proofplan]

[step:Find a point outside any prescribed pair of distinct lines]Let $\ell,m \in L$ be distinct lines. We prove that there exists $x \in P$ such that $x \notin P(\ell) \cup P(m)$. By the four-point axiom, choose points $a,b,c,d \in P$ such that no three of them are incident with a common line. If one of these four points is not in $P(\ell) \cup P(m)$, we are done. Suppose instead that \begin{align*} \{a,b,c,d\} \subset P(\ell) \cup P(m). \end{align*} Since no three of $a,b,c,d$ lie on one line, each of $\ell$ and $m$ contains at most two of these four points: \begin{align*} |\{a,b,c,d\}\cap P(\ell)| \leq 2, \qquad |\{a,b,c,d\}\cap P(m)| \leq 2. \end{align*} Because all four points are assumed to lie in $P(\ell) \cup P(m)$, both inequalities must be equalities and no chosen point can lie in $P(\ell)\cap P(m)$; otherwise the union would contain at most three of the chosen points. Hence, after relabelling, we may assume \begin{align*} a,b \in P(\ell) \end{align*} and \begin{align*} c,d \in P(m) \setminus P(\ell). \end{align*} Let $r \in L$ be the unique line incident with $a$ and $c$, and let $s \in L$ be the unique line incident with $b$ and $d$. The lines $r$ and $s$ are distinct: if $r=s$, then $a,b,c$ would all be incident with $r$, contradicting the four-point axiom. Therefore the axiom that two distinct lines meet in a unique point applies to $r$ and $s$; let $x \in P$ be the unique point incident with both $r$ and $s$. We claim that $x \notin P(\ell)$. If $x \in P(\ell)$ and $x \neq a$, then the two distinct points $a$ and $x$ lie on both $r$ and $\ell$, so $r=\ell$ by uniqueness of the line through two points. This would imply $c \in P(\ell)$, contradicting that $a,b,c$ are not collinear. If $x=a$, then $a$ lies on $s$, so $a,b,d$ all lie on $s$ because $b,d \in P(s)$; this again contradicts the four-point choice. Therefore $x \notin P(\ell)$. The same argument with $(\ell,a,b)$ replaced by $(m,c,d)$ gives $x \notin P(m)$. Thus $x \notin P(\ell) \cup P(m)$.[/step]

[guided]We need a point outside two given lines $\ell$ and $m$. The four-point axiom supplies four points $a,b,c,d$ with no three collinear. If any one of them already avoids both $\ell$ and $m$, there is nothing more to prove. So assume all four points lie on $P(\ell) \cup P(m)$. Since no three of them may lie on a single line, $\ell$ contains at most two of the four and $m$ contains at most two of the four: \begin{align*} |\{a,b,c,d\}\cap P(\ell)| \leq 2, \qquad |\{a,b,c,d\}\cap P(m)| \leq 2. \end{align*} Because $P(\ell)\cup P(m)$ contains all four chosen points, these two upper bounds force exactly two chosen points on $\ell$ and exactly two chosen points on $m$. They also force no chosen point to lie in both $P(\ell)$ and $P(m)$; if one did, the two sets would cover at most three of the chosen points. Relabel them so that \begin{align*} a,b \in P(\ell) \end{align*} and \begin{align*} c,d \in P(m) \setminus P(\ell). \end{align*} Now form the two diagonal lines. Let $r \in L$ be the unique line through $a$ and $c$, and let $s \in L$ be the unique line through $b$ and $d$. Before using the axiom that two distinct lines meet, we verify distinctness. If $r=s$, then $a,b,c$ would all lie on $r$, contradicting that no three of $a,b,c,d$ are collinear. Hence $r\neq s$, and the line-intersection axiom gives a unique point $x \in P$ incident with both $r$ and $s$. We verify that this diagonal intersection $x$ cannot lie on $\ell$. If $x \in P(\ell)$ and $x \neq a$, then the two distinct points $a$ and $x$ are incident with both $r$ and $\ell$. The uniqueness axiom for the line through two distinct points forces $r=\ell$. But then $c \in P(r)=P(\ell)$, so $a,b,c$ are three collinear points, contradicting the four-point axiom. If instead $x=a$, then $a$ lies on $s$, and since $b,d \in P(s)$, the three points $a,b,d$ lie on $s$, again a contradiction. Hence $x \notin P(\ell)$. The same verification for the line $m$ shows $x \notin P(m)$. Indeed, if $x \in P(m)$ and $x \neq c$, then $r=m$, forcing $a,c,d$ to be collinear; if $x=c$, then $c,b,d$ lie on $s$. Both alternatives contradict the choice of $a,b,c,d$. Therefore $x$ lies outside $P(\ell) \cup P(m)$.[/guided]

[step:Compare a pencil through an external point with the points on a line] Let $p \in P$ and $\ell \in L$ satisfy $p \notin P(\ell)$. Define $\Phi_{p,\ell}: L(p) \to P(\ell)$ as follows. For each $m \in L(p)$, let $\Phi_{p,\ell}(m)$ be the unique point in $P(m)\cap P(\ell)$. This map is well-defined by the axiom that two distinct lines meet in a unique point; also $m \neq \ell$ because $p \in P(m)$ but $p \notin P(\ell)$. The map $\Phi_{p,\ell}$ is injective. If $m_1,m_2 \in L(p)$ and $\Phi_{p,\ell}(m_1)=\Phi_{p,\ell}(m_2)=q$, then both $m_1$ and $m_2$ contain the two distinct points $p$ and $q$. The uniqueness axiom for the line through two distinct points gives $m_1=m_2$. The map $\Phi_{p,\ell}$ is surjective. If $q \in P(\ell)$, then $q \neq p$ because $p \notin P(\ell)$. The unique line through $p$ and $q$ lies in $L(p)$ and maps to $q$. Hence $\Phi_{p,\ell}$ is a bijection, so \begin{align*} |L(p)| = |P(\ell)|. \end{align*} [/step]

[step:Show that all lines contain the same number of points] Let $\ell,m \in L$ be distinct lines. By the first step, choose $x \in P$ such that \begin{align*} x \notin P(\ell) \cup P(m). \end{align*} Applying the bijection from the previous step to the pair $(x,\ell)$ gives \begin{align*} |L(x)| = |P(\ell)|. \end{align*} Applying it to the pair $(x,m)$ gives \begin{align*} |L(x)| = |P(m)|. \end{align*} Therefore $|P(\ell)|=|P(m)|$. Since $\ell$ and $m$ were arbitrary distinct lines, all lines contain a common number of points. Denote this common number by $n+1$. It remains to prove $n \geq 2$. Choose points $a,b,c,d \in P$ such that no three of them are incident with a common line. Let $r \in L$ be the unique line incident with $a$ and $b$, and let $s \in L$ be the unique line incident with $c$ and $d$. The lines $r$ and $s$ are distinct; if $r=s$, then $a,b,c$ would be three of the chosen points on one line. Let $y \in P$ be the unique point incident with both $r$ and $s$. The point $y$ is distinct from $a$ and $b$, because $y=a$ or $y=b$ would put that point on $s$ and hence make three of $a,b,c,d$ collinear with $c,d$. Therefore the line $r$ contains the three distinct points $a,b,y$. Since all lines have the common size $n+1$, every line contains at least three points. Thus $n+1 \geq 3$, and hence $n \geq 2$. [/step]

[step:Show that every point lies on the same number of lines] Let $p \in P$. We first prove that there is at least one line $\ell \in L$ with $p \notin P(\ell)$. By the four-point axiom, choose points $a,b,c,d \in P$ such that no three of them are incident with a common line. Choose $u \in \{a,b,c,d\}$ with $u \neq p$, and choose two distinct points $v,w \in \{a,b,c,d\}\setminus\{u,p\}$; this is possible because the four-point set has four elements. Let $\ell_v \in L$ be the unique line incident with $u$ and $v$, and let $\ell_w \in L$ be the unique line incident with $u$ and $w$. If both $p \in P(\ell_v)$ and $p \in P(\ell_w)$ held, then the two distinct points $u$ and $p$ would lie on both $\ell_v$ and $\ell_w$, so uniqueness of the line through two distinct points would give $\ell_v=\ell_w$. Then $u,v,w$ would be three of the four chosen points on one line, contradicting the four-point axiom. Hence at least one of $\ell_v$ and $\ell_w$ is not incident with $p$; denote such a line by $\ell$. Applying the pencil-line bijection to $(p,\ell)$ gives \begin{align*} |L(p)| = |P(\ell)| = n+1. \end{align*} Thus every point lies on exactly $n+1$ lines. [/step]

[step:Count all points from the lines through one point] Fix $p \in P$. Since $|L(p)|=n+1$, list the lines through $p$ as \begin{align*} L(p)=\{\ell_1,\dots,\ell_{n+1}\}. \end{align*} Every point $q \in P$ with $q \neq p$ lies on exactly one of these lines: namely, the unique line through $p$ and $q$. Distinct lines through $p$ have no common point other than $p$, because two distinct common points would violate uniqueness of the line through two points. Each line $\ell_i$ contains $n+1$ points, one of which is $p$, so it contributes exactly $n$ points distinct from $p$. Therefore The preceding disjoint decomposition gives \begin{align*} |P| = 1 + \sum_{i=1}^{n+1} |P(\ell_i)\setminus\{p\}|. \end{align*} For each $i \in \{1,\dots,n+1\}$, the set $P(\ell_i)\setminus\{p\}$ has cardinality $n$, so \begin{align*} |P| = 1 + \sum_{i=1}^{n+1} n. \end{align*} Evaluating the finite sum gives \begin{align*} |P| = 1+n(n+1). \end{align*} Thus \begin{align*} |P| = n^2+n+1. \end{align*} [/step]

[step:Count incidences to obtain the number of lines] Let \begin{align*} \mathcal I := \{(p,\ell)\in P\times L : (p,\ell)\in I\} \end{align*} be the incidence set. Counting $\mathcal I$ by points gives \begin{align*} |\mathcal I|=\sum_{p\in P}|L(p)|=(n+1)|P|. \end{align*} Counting $\mathcal I$ by lines gives \begin{align*} |\mathcal I|=\sum_{\ell\in L}|P(\ell)|=(n+1)|L|. \end{align*} Since $n+1>0$, these two equalities imply $|P|=|L|$. From the previous step, \begin{align*} |L|=|P|=n^2+n+1. \end{align*} This proves all asserted parameter formulas. [/step]