Solution:
Note first that the triple $(0, 0, 0)$ is a solution of the system.

If $y = 0$, then it follows from the first equation that $x = -z$ and the second one gives that $x = z = 0$.

If $y \neq 0$, set $a = \frac{x}{y}$ and $b = \frac{z}{y}$. Then the system becomes
$$\left\lvert\, \begin{aligned} & 1 + a + b = 3 a y \\ & 1 + a^{2} + b^{2} = 3 a b \\ & y\left(1 + a^{3} + b^{3}\right) = 3 b \end{aligned} \right.$$
Hence $y = \frac{1 + a + b}{3 a}$ and therefore
$$\left\lvert\, \begin{aligned} & (1 + a + b)\left(1 + a^{3} + b^{3}\right) = 9 a b \\ & 1 + a^{2} + b^{2} = 3 a b \end{aligned} \right.$$
Set $u = a + b$ and $v = a b$. Then
$$\left\lvert\, \begin{aligned} & (1 + u)\left(1 + u^{3} - 3 u v\right) = 9 v \\ & 1 + u^{2} - 2 v = 3 v \end{aligned} \right.$$
Hence $v = \frac{u^{2} + 1}{5}$ and plugging it into the first equation gives
$$0 = u^{4} + u^{3} - 6 u^{2} + u - 2 = (u - 2)\left(u^{3} + 3 u^{2} + 1\right)$$
The case $u = 2$ leads to $v = 1$, $a = b = 1$ and gives the solution $(x, y, z) = (1, 1, 1)$.

The function $f(u) = u^{3} + 3 u^{2} + 1$ has a local maximum at $u = -2$ and a local minimum at $u = 0$. Since $f(0) = 1 > 0$, the equation $f(u) = 0$ has only one real root $u_{0}$ and $u_{0} < -2$. Then $u_{0}^{2} - 4 \frac{u_{0}^{2} + 1}{5} = \frac{u_{0}^{2} - 4}{5} > 0$ which shows that the system
$$\left\lvert\, \begin{aligned} & a + b = u_{0} \\ & a b = \frac{u_{0}^{2} + 1}{5} \end{aligned} \right.$$
has two solutions.

Thus the given system has four real solutions.