Squares, square, everywhere!

We take $\pmod{4}$ of both sides of equation. Since any perfect square can only be equal to $1,0 \pmod {4}$ , then we cannot say that $x,y,z$ are odd, since we get that $3=\pm 2$ when taken into account $\pmod 4$ , which is absurd.

WLOG, suppose only 1 $x$ is even . Then we get that $x^2+y^2+z^2 \equiv y^2+z^2 \pmod 4$ , but since $2xyz \equiv 0 \pmod 4$ , then this implies that the RHS must also be $0 \pmod 4$ , but the only way to achieve this is when $y, z$ are even.

This implies that $x=2x_1, y=2y_1, z=2z_1$ for some integers $x_1, y_1, z_1$ . We substitute this into the original equation to get a new equation $4x_1^2+4y_1^2+4z_1^2 = 16x_1y_1z_1$ , which simplifies to $x_1^2+y_1^2+z_1^2 = 4x_1y_1z_1$ Repeating the same logic as above, we get that $x_1, y_1, z_1$ must also be even, i.e. $x_1=2x_2, y_1=2y_2, z_1=2z_2$ , and we can repeat this process ad infinitum. This implies that $\frac{x}{2^n}$ must be an integer for all positive integral values of $n$ , and the only integer which satisfies this is $x=0$ . So, if $x=0$ , then $xyz = \boxed{0}$ , which is our final answer.

If you didn't get it, the answer explains itself.