This is a system of three cubic equations. We perform elimination and reduction to solve it.
Obviously, when $y=0$, from (1) we get $x=-z$, and then from (2) we derive $x=z=0$. Thus, the triplet $(0,0,0)$ satisfies the system. Suppose $y \neq 0$, let $a=\frac{x}{y}, b=\frac{z}{y}$. Therefore, the original system transforms into
\[
\left\{\begin{array}{l}
1+a+b=3 a y, \\
1+a^{2}+b^{2}=3 a b, \\
y\left(1+a^{3}+b^{3}\right)=3 b .
\end{array}\right.
\]
By substituting $\frac{1+a+b}{3 a}$ for $y$, we get
\[
\left\{\begin{array}{l}
(1+a+b)\left(1+a^{3}+b^{3}\right)=9 a b, \\
1+a^{2}+b^{2}=3 a b .
\end{array}\right.
\]

Let $u=a+b, v=a b$, then we have
\[
\left\{\begin{array}{l}
(1+u)\left(1+u^{3}-3 u v\right)=9 v, \\
1+u^{2}-2 v=3 v .
\end{array}\right.
\]

Thus, $v=\frac{u^{2}+1}{5}$. Therefore,
\[
0=u^{4}+u^{3}-6 u^{2}+u-2=(u-2)\left(u^{3}+3 u^{2}+1\right).
\]

When $u=2$, we get $v=1$, which means $a=b=1$.
Thus, we obtain one solution $(x, y, z)=(1,1,1)$.
Next, consider the function $f(u)=u^{3}+3 u^{2}+1$. When $u=-2$, it reaches a local maximum; when $u=0$, it reaches a local minimum. Since $f(0)=1>0$, the equation $f(u)=0$ has a unique real root $u_{0}$, where $u_{0}<0$. Therefore,
the system \(\left\{\begin{array}{l}a+b=u_{0} \\ a b=\frac{u_{0}^{2}+1}{5}\end{array}\right.\) has two solutions.
Thus, we have obtained two additional solutions for the given system. Hence, the given system has 4 sets of real solutions.
Example 9 (2003 Bulgarian National Mathematical Olympiad) Find the number of real solutions of the system of equations $\left\{\begin{array}{l}x+y+z=3 x y, \\ x^{2}+y^{2}+z^{2}=3 x z, \\ x^{3}+y^{3}+z^{3}=3 y z\end{array}\right.$.