General Diophantine

Let $(x,y,z)$ be a solution to the equation with $z\geq 1$ minimal over all such solutions. Then $x^2 + y^2 = 4^z \equiv 0 \pmod{4}.$ However, $a^2 \equiv 0,1 \pmod{4}$ depending on whether $a$ is even or odd, respectively, so in particular, if at least one of $x,y$ is odd, then $x^2 + y^2 \equiv 1,2 \pmod{4}$ so we see both $x,y$ are even. That is, there are positive integers $s,t$ such that $x=2s, y=2t$ . Substituting this into the original equation, we have $4s^2 + 4t^2 = (2s)^2 + (2t)^2 = x^2 + y^2 = 4^z,$ and therefore $s^2 + t^2 = 4^{z-1}.$

Now, $(s,t,z-1)$ satisfies the original equation and $s,t$ are positive integers. If additionally $z-1 < z$ were a positive integer, then it would contradict the initial choice of $z$ as the minimal such positive integer. Therefore $z-1$ cannot be a positive integer, and since $z$ was a positive integer, this forces $z-1=0$ . We use this to rewrite the equation involving $s$ and $t$ as $s^2 + t^2 = 4^0 = 1.$

This however, is impossible. If $s,t$ are nonzero integers, then $s^2,t^2 \geq 1$ , so their sum cannot be $1$ .

Since assuming we had a solution led to a contradiction, we may conclude there are no solutions, and the answer is $0$ .