Solution:
We have
$$\begin{aligned} & x^{3} y z = z^{2} - 2 y^{2} \\ & y^{3} z x = x^{2} - 2 z^{2} \\ & z^{3} x y = y^{2} - 2 x^{2} \end{aligned}$$
with $x y z \neq 0$.
Adding these up we obtain $\left(x^{2} + y^{2} + z^{2}\right)(x y z + 1) = 0$. Hence $x y z = -1$. Now the system of equations becomes:
$$\begin{aligned} & x^{2} = 2 y^{2} - z^{2} \\ & y^{2} = 2 z^{2} - x^{2} \\ & z^{2} = 2 x^{2} - y^{2} \end{aligned}$$
Then the first two equations give $x^{2} = y^{2} = z^{2}$. As $x y z = -1$, we conclude that $(x, y, z) = (1, 1, -1), (1, -1, 1), (-1, 1, 1)$ and $(-1, -1, -1)$ are the only solutions.