Theorem 1   $x^4+y^4 =z^2$ has no solution in positive integers.

Proof. Suppose we had a positive $z$ such that $x^4+y^4=z^2$ holds. We may assume $\gcd(x,y,z)=1$ . Then $z$ must be odd, and $x,y$ have opposite parity. Since $(x^2)^2 +(y^2)^2 =z^2$ is a primitive Pythagorean triple, we have \begin{equation} \label{eq1} x^2=2pq, y^2 =q^2-p^2, z=p^2+q^2 \end{equation}where $p,q \in \N$ , $p<q$ are coprime and have opposite parity. Since $y^2+p^2=q^2$ is a primitive Pythagorean triple, we have coprime $s,r \in \N$ , $s<r$ of opposite parity satisfying \begin{equation} \label{eq2} q=r^2+s^2, y=r^2-s^2, p=2rs. \end{equation}From $\gcd(r^2, s^2)=1$ it follows that $\gcd(r^2, r^2+s^2)=1=\gcd(s^2, r^2+s^2)$ , which implies $\gcd(rs, r^2+s^2)=1$ . Since $\left(\frac{x}{2}\right)^2 = \frac{pq}{2} = rs(r^2+s^2)$ is a square, each of $r,s,r^2+s^2$ is a square.

Setting $Z^2 =q$ , $X^2 =r$ , $Y^2=s$ $q=r^2+s^2$ leads to \begin{equation} \label{eq3} Z^2=X^4+Y^4 \end{equation}where $Z^2=q<q^2+p^2=z <z^2$ . Thus, equation gives a solution where $Z< z$ . Applying the above steps repeatedly would produce an infinite sequence $z > Z > z_2 > \ldots$ of positive integers, each of which was the sum of two fourth powers. But there cannot be infinitely many positive integers smaller than a given one; in particular this contradicts to the fact that there must exist a smallest $z$ for which () is solvable. So there are no solutions in positive integers for this equation.