 Fourth-power sum cannot be a square

Choose a primitive counterexample with minimal $z$ and $y$ even. Its Pythagorean triple $(x^2,y^2,z)$ has $x^2=m^2-n^2$, $y^2=2mn$. Modulo four forces $m$ odd and $n$ even, so $m=u^2$ and $n=2v^2$. The primitive triangle $(x,2v^2,u^2)$ then has a parametrization with $v^2=rs$ and $u^2=r^2+s^2$. Coprimality makes $r=a^2,s=b^2$, giving $a^4+b^4=u^2$ with $u<z$. This contradicts infinite descent.