[proofplan] We prove the stronger assertion by infinite descent on the positive integer $z$. Starting from a solution with $z$ minimal, we first reduce to a primitive solution and regard $(x^2,y^2,z)$ as a primitive Pythagorean triple. The standard parametrisation of primitive Pythagorean triples, together with repeated coprime-factor arguments, forces several products of coprime integers to be squares. These square decompositions produce a new solution $r^4+s^4=b^2$ with $0<b<z$, contradicting the minimality of $z$. [/proofplan]

[step:Record the elementary arithmetic facts used in the descent] We shall use the following two elementary facts. [claim:Coprime factors of a square are squares] If $a,b \in \mathbb{Z}_{>0}$ satisfy $\gcd(a,b)=1$ and $ab$ is a square in $\mathbb{Z}$, then there exist $u,v \in \mathbb{Z}_{>0}$ such that $a=u^2$ and $b=v^2$. [/claim] [proof] Write the prime factorizations \begin{align*} a=\prod_{p} p^{\alpha_p}, \qquad b=\prod_{p} p^{\beta_p}, \end{align*} where all but finitely many exponents $\alpha_p,\beta_p \in \mathbb{Z}_{\geq 0}$ are zero. Since $\gcd(a,b)=1$, for each prime $p$ at most one of $\alpha_p,\beta_p$ is nonzero. Since $ab$ is a square, every exponent $\alpha_p+\beta_p$ is even. Hence every nonzero $\alpha_p$ and every nonzero $\beta_p$ is even. Therefore $a$ and $b$ are squares. [/proof] [claim:Primitive Pythagorean triples have the Euclidean form] Let $A,B,C \in \mathbb{Z}_{>0}$ satisfy \begin{align*} A^2+B^2=C^2, \qquad \gcd(A,B)=1, \end{align*} and suppose $B$ is even. Then there exist coprime integers $m,n \in \mathbb{Z}_{>0}$ with $m>n$, $m$ and $n$ of opposite parity, such that \begin{align*} A=m^2-n^2, \qquad B=2mn, \qquad C=m^2+n^2. \end{align*} [/claim] [proof] Since $\gcd(A,B)=1$ and $B$ is even, $A$ is odd. Hence $C$ is odd as well. From \begin{align*} B^2=C^2-A^2=(C-A)(C+A), \end{align*} both $C-A$ and $C+A$ are positive even integers. Define \begin{align*} R:=\frac{C+A}{2}, \qquad S:=\frac{C-A}{2}. \end{align*} Then $R,S \in \mathbb{Z}_{>0}$, $R>S$, and \begin{align*} RS=\frac{(C+A)(C-A)}{4}=\left(\frac{B}{2}\right)^2. \end{align*} We claim $\gcd(R,S)=1$. If a prime $p$ divides both $R$ and $S$, then $p$ divides $R+S=C$ and $R-S=A$. Hence $p$ divides $C^2-A^2=B^2$, so $p$ divides $B$. Thus $p$ divides $A$ and $B$, contradicting $\gcd(A,B)=1$. Therefore $\gcd(R,S)=1$. By the preceding claim, there exist $m,n \in \mathbb{Z}_{>0}$ such that \begin{align*} R=m^2, \qquad S=n^2. \end{align*} Thus \begin{align*} C=m^2+n^2, \qquad A=m^2-n^2, \qquad B^2=4m^2n^2. \end{align*} Since $B>0$, we get $B=2mn$. The equality $\gcd(R,S)=1$ gives $\gcd(m,n)=1$. Finally, $m$ and $n$ cannot both be odd, because then $m^2-n^2$ would be even, contradicting that $A$ is odd. Since they are coprime, they cannot both be even. Hence $m$ and $n$ have opposite parity. [/proof] [/step]

[step:Choose a primitive minimal counterexample] Assume, for contradiction, that there exist positive integers $x,y,z \in \mathbb{Z}$ satisfying \begin{align*} x^4+y^4=z^2. \end{align*} Among all such triples, choose one with $z$ minimal. We first show that $\gcd(x,y)=1$. Let $g:=\gcd(x,y)$. If $g>1$, write $x=gx_1$ and $y=gy_1$ with $x_1,y_1 \in \mathbb{Z}_{>0}$. Then \begin{align*} g^4(x_1^4+y_1^4)=z^2. \end{align*} Thus $g^2$ divides $z$, so $z=g^2z_1$ for some $z_1 \in \mathbb{Z}_{>0}$. Dividing by $g^4$ gives \begin{align*} x_1^4+y_1^4=z_1^2. \end{align*} Since $g>1$, we have $z_1<z$, contradicting the minimality of $z$. Therefore $\gcd(x,y)=1$. The fourth power of an odd integer is congruent to $1$ modulo $16$, while the fourth power of an even integer is congruent to $0$ modulo $16$. Since a square is congruent only to $0,1,4,$ or $9$ modulo $16$, the integers $x$ and $y$ cannot both be odd. They also cannot both be even because $\gcd(x,y)=1$. Hence exactly one of $x,y$ is even. Since the equation is symmetric in $x$ and $y$, interchange them if necessary and assume \begin{align*} x \text{ is even}, \qquad y \text{ is odd}. \end{align*} [/step]

[step:Parametrize the primitive Pythagorean triple $(y^2,x^2,z)$] The equality \begin{align*} (y^2)^2+(x^2)^2=z^2 \end{align*} shows that $(y^2,x^2,z)$ is a Pythagorean triple. Since $\gcd(x,y)=1$, we also have $\gcd(y^2,x^2)=1$, and $x^2$ is even. Applying the primitive Pythagorean parametrisation with \begin{align*} A:=y^2, \qquad B:=x^2, \qquad C:=z, \end{align*} there exist coprime integers $m,n \in \mathbb{Z}_{>0}$ with $m>n$ and opposite parity such that \begin{align*} y^2=m^2-n^2, \qquad x^2=2mn, \qquad z=m^2+n^2. \end{align*} [/step]

[step:Show that the even parameter is $n$ and extract square factors from $2mn$] We claim that $m$ is odd and $n$ is even. Since $m$ and $n$ have opposite parity, it suffices to rule out the case in which $m$ is even. Suppose $m$ is even. Since $\gcd(m,n)=1$ and \begin{align*} x^2=2mn, \end{align*} prime factorization gives positive integers $a,b \in \mathbb{Z}_{>0}$ such that \begin{align*} m=2a^2, \qquad n=b^2. \end{align*} Then \begin{align*} y^2=m^2-n^2=(m-n)(m+n). \end{align*} The integers $m-n$ and $m+n$ are positive and odd. Their greatest common divisor divides both their sum $2m$ and their difference $2n$; since it is odd and $\gcd(m,n)=1$, it equals $1$. Hence, by the coprime-square factor claim, there exist $c,d \in \mathbb{Z}_{>0}$ such that \begin{align*} m-n=c^2, \qquad m+n=d^2. \end{align*} Adding these equations gives \begin{align*} c^2+d^2=2m=4a^2. \end{align*} But $c$ and $d$ are odd, so $c^2+d^2 \equiv 1+1 \equiv 2 \pmod 4$, contradicting $4a^2 \equiv 0 \pmod 4$. Therefore $m$ is odd and $n$ is even. Because $\gcd(m,n)=1$ and $x^2=2mn$, the prime factorization of $2mn$ now forces \begin{align*} m=b^2, \qquad n=2a^2 \end{align*} for some coprime positive integers $a,b \in \mathbb{Z}_{>0}$. [/step]

[step:Split $y^2=(b^2-2a^2)(b^2+2a^2)$ into coprime squares] Substituting $m=b^2$ and $n=2a^2$ into $y^2=m^2-n^2$ gives \begin{align*} y^2=(b^2-2a^2)(b^2+2a^2). \end{align*} Both factors are positive, because $m>n$. They are odd, since $b$ is odd. Let \begin{align*} D:=\gcd(b^2-2a^2,\ b^2+2a^2). \end{align*} Then $D$ divides their sum $2b^2$ and their difference $4a^2$. Since $D$ is odd and $\gcd(a,b)=1$, we have $D=1$. Therefore the two factors are coprime positive integers whose product is a square. By the coprime-square factor claim, there exist $c,d \in \mathbb{Z}_{>0}$ such that \begin{align*} b^2-2a^2=c^2, \qquad b^2+2a^2=d^2. \end{align*} Subtracting the first identity from the second gives \begin{align*} d^2-c^2=4a^2. \end{align*} [/step]

[step:Construct a smaller fourth-power solution] Since $c$ and $d$ are odd and $d>c$, define \begin{align*} e:=\frac{d-c}{2}, \qquad f:=\frac{d+c}{2}. \end{align*} Then $e,f \in \mathbb{Z}_{>0}$ and \begin{align*} ef=\frac{d^2-c^2}{4}=a^2. \end{align*} Moreover, $\gcd(e,f)=1$: any common divisor of $e$ and $f$ divides $e+f=d$ and $f-e=c$, and $\gcd(c,d)=1$ because $c^2$ and $d^2$ arose from coprime factors. Hence, by the coprime-square factor claim, there exist $r,s \in \mathbb{Z}_{>0}$ such that \begin{align*} e=r^2, \qquad f=s^2. \end{align*} Thus \begin{align*} d=e+f=r^2+s^2, \qquad c=f-e=s^2-r^2. \end{align*} Using these formulas, compute \begin{align*} b^2 &=\frac{(b^2-2a^2)+(b^2+2a^2)}{2} \\ &=\frac{c^2+d^2}{2} \\ &=\frac{(s^2-r^2)^2+(r^2+s^2)^2}{2} \\ &=r^4+s^4. \end{align*} Therefore $(r,s,b)$ is another positive integer solution of the same type: \begin{align*} r^4+s^4=b^2. \end{align*} But the original hypotenuse satisfies \begin{align*} z=m^2+n^2=b^4+4a^4. \end{align*} Since $a,b \geq 1$, we have $b<z$. This contradicts the minimal choice of $z$. [/step]

[step:Conclude the nonexistence of both equations] The contradiction shows that no positive integers $x,y,z \in \mathbb{Z}$ satisfy \begin{align*} x^4+y^4=z^2. \end{align*} If positive integers $x,y,w \in \mathbb{Z}$ satisfied \begin{align*} x^4+y^4=w^4, \end{align*} then setting $z:=w^2$ would give \begin{align*} x^4+y^4=z^2, \end{align*} contradicting the result just proved. Hence there are no positive integer solutions to \begin{align*} x^4+y^4=w^4. \end{align*} [/step]