[proofplan]
We construct the correspondence in both directions. A primitive integer triple gives a rational projective point because the Pythagorean equation is exactly the homogeneous conic equation. Conversely, every rational point has rational coordinates, and clearing denominators gives an integer solution; dividing by the greatest common divisor makes it primitive without changing the projective point. Finally, projective equality between primitive integer representatives forces the scalar relating them to be $\pm 1$, which is exactly the stated sign ambiguity.
[/proofplan]
[step:Verify that primitive triples define rational points on the conic]
Let $[(x,y,z)] \in T/{\sim}$ be an equivalence class, with representative $(x,y,z) \in T$. Since $\gcd(x,y,z)=1$, the triple $(x,y,z)$ is not $(0,0,0)$. Thus $[x:y:z]$ is a point of $\mathbb{P}^2(\mathbb{Q})$. Since $(x,y,z) \in T$, we have
\begin{align*}
x^2+y^2=z^2,
\end{align*}
so $[x:y:z] \in C(\mathbb{Q})$.
The definition is independent of the representative of the class in $T/{\sim}$: if $(x',y',z')=(-x,-y,-z)$, then
\begin{align*}
[x':y':z']=[-x:-y:-z]=[x:y:z]
\end{align*}
in $\mathbb{P}^2(\mathbb{Q})$, because multiplication by $-1 \in \mathbb{Q}^\times$ does not change a projective point. Hence $\Phi$ is well-defined.
[/step]
[step:Clear denominators and divide by the greatest common divisor]
Let $[X:Y:Z] \in C(\mathbb{Q})$. Choose a representative $(X,Y,Z) \in \mathbb{Q}^3 \setminus \{(0,0,0)\}$ satisfying
\begin{align*}
X^2+Y^2=Z^2.
\end{align*}
Choose an integer $d \in \mathbb{Z}$ with $d \neq 0$ such that
\begin{align*}
a := dX,\qquad b := dY,\qquad c := dZ
\end{align*}
are all integers. Then $(a,b,c) \in \mathbb{Z}^3 \setminus \{(0,0,0)\}$ and, multiplying the conic equation by $d^2$,
\begin{align*}
a^2+b^2=c^2.
\end{align*}
Let
\begin{align*}
g := \gcd(a,b,c).
\end{align*}
Since $(a,b,c)$ is nonzero, $g \in \mathbb{N}$. Define
\begin{align*}
x := \frac{a}{g},\qquad y := \frac{b}{g},\qquad z := \frac{c}{g}.
\end{align*}
Then $(x,y,z) \in \mathbb{Z}^3$, $\gcd(x,y,z)=1$, and division of the equation $a^2+b^2=c^2$ by $g^2$ gives
\begin{align*}
x^2+y^2=z^2.
\end{align*}
Thus $(x,y,z) \in T$. Moreover,
\begin{align*}
[x:y:z]=[a:b:c]=[dX:dY:dZ]=[X:Y:Z],
\end{align*}
so every point of $C(\mathbb{Q})$ lies in the image of $\Phi$.
[/step]
[step:Show that primitive representatives differ only by sign]
Suppose $(x,y,z),(x',y',z') \in T$ satisfy
\begin{align*}
[x:y:z]=[x':y':z'].
\end{align*}
By equality in $\mathbb{P}^2(\mathbb{Q})$, there exists $\lambda \in \mathbb{Q}^\times$ such that
\begin{align*}
(x',y',z')=(\lambda x,\lambda y,\lambda z).
\end{align*}
Write $\lambda=m/n$ with $m,n \in \mathbb{Z}$, $n>0$, and $\gcd(m,n)=1$.
Since $x'=\lambda x$, $y'=\lambda y$, and $z'=\lambda z$ are integers, the integer $n$ divides each of $mx$, $my$, and $mz$. Because $\gcd(m,n)=1$, Euclid's lemma gives
\begin{align*}
n \mid x,\qquad n \mid y,\qquad n \mid z.
\end{align*}
Since $\gcd(x,y,z)=1$, this forces $n=1$. Therefore $\lambda=m$ is an integer.
Applying the same argument to
\begin{align*}
(x,y,z)=\lambda^{-1}(x',y',z')
\end{align*}
shows that $\lambda^{-1}$ is also an integer. Hence $\lambda \in \mathbb{Z}$ and $\lambda^{-1} \in \mathbb{Z}$, so $\lambda=\pm 1$. Therefore
\begin{align*}
(x',y',z')=(x,y,z)
\end{align*}
or
\begin{align*}
(x',y',z')=(-x,-y,-z).
\end{align*}
Thus the two triples represent the same class in $T/{\sim}$, proving that $\Phi$ is injective.
[/step]
[step:Conclude the bijection]
The first step shows that $\Phi$ is a well-defined map from $T/{\sim}$ to $C(\mathbb{Q})$. The denominator-clearing step shows that every rational point of the conic is in its image, so $\Phi$ is surjective. The primitive-representative step shows that two primitive triples mapping to the same rational projective point differ exactly by the allowed simultaneous sign change, so $\Phi$ is injective. Hence $\Phi$ is a bijection.
[/step]
