The only solution is $\boxed{x=y=z=0}$. To see this, we note that $x$, $y$ and $z$ cannot all be odd because otherwise the value of $(x^2 + y^2 + z^2 -2xyz)$ will be odd and therefore cannot be zero. Therefore, at least one of $x$, $y$, and $z$ must be even. It follows that $$2 \mid xyz \implies 4 \mid (2xyz)$$ It follows that both sides of the following equation is a multiple of $4$: $$x^2 + y^2 + z^2=2xyz$$ Because a square of an odd integer must be congruent to $1$ modulo $4$, the left side $(x^2+y^2+z^2)$ can be a multiple of $4$ only when all of the $x$, $y$ and $z$ are even. Let $x=2x_1$, $y=2y_1$, and $z=2z_1$. Then we have $$4x_1^2 + 4y_1^2 + 4z_1^2 = 16x_1y_1z_1\implies x_1^2 + y_1^2 + z_1^2 = 4x_1y_1z_1$$ By the same reasoning, we find $x_1$, $y_1$, and $z_1$ are all even. Letting $x_1=2x_2$, $y_1=2y_2$, and $z_1=2z_2$ will lead to $$x_2^2 + y_2^2 + z_2^2 = 8x_2y_2z_2$$ Again, we find $x_2$, $y_2$, and $z_2$ are even. This process can continue infinitely which is impossible by the principle of infinite descent, unless $x=y=z=0$. Therefore, we conclude $(0, 0, 0)$ is the only solution.