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

where $p,q\in{\mathbb{N}}$, $p<q$ are coprime and have opposite parity. Since $y^{2}+p^{2}=q^{2}$ is a primitive Pythagorean triple, we have coprime $s,r\in{\mathbb{N}}$, $s<r$ of opposite parity satisfying

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

where $Z^{2}=q<q^{2}+p^{2}=z<z^{2}$. Thus, equation 3 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 (1) is solvable. So there are no solutions in positive integers for this equation. ∎