If this equation has positive integer solution, let $(x_0, y_0, z_0)$ be the solution that minimizes the value of $z$.

First, let's show that $x_0$ and $y_0$ are relatively prime. If this is not true, let $\gcd(x_0, y_0)= d > 1$. It follows that $d^4 \mid x_0^4 + y_0^4 = z_0^2$. Hence $d^2 \mid z_0$. Therefore $(\frac{x_0}{d}, \frac{y_0}{d}, \frac{x_0}{d^2})$ is another positive integer solution with smaller $z$. This is a contradiction. If $(x_0, y_0)$ are relative prime, then $(x_0^2, y_0^2, z_0)$ is a primitive Pythagorean triplet.

Let \begin{equation} \left\{ \begin{aligned} x^2 &= m^2 - n^2 \\ y^2 &= 2mn \\ z &=m^2 + n^2 \end{aligned} \right. \end{equation} where positive integers $m$ and $n$ are relatively prime. From this, we have another Pythagorean triplet: $x^2 + n^2 = m^2$. Hence, there exist positive integers $a$ and $b$ such that: \begin{equation} \left\{ \begin{aligned} x &= a^2 - b^2 \\ n &= 2ab \\ m &= a^2 + b^2 \end{aligned} \right. \end{equation} Combining these two systems leads to $$y^2 = 2mn = 4ab(a^2+b^2)$$ Because $a$ and $b$ are relatively prime, it can be shown that $ab$ and $(a^2 + b^2)$ are relatively prime too. Consequently both of them must be square numbers. That is, there exits a positive integer $k$ such that $k^2 = a^2 + b^2 = m$. However, this means we found another $a^2 + b^2 = m$ that is less than $m^2 + n^2 = z$. By the principle of infinite descent, this is impossible.