Example 8 Find the real number pairs $(x, y, z)$ that satisfy the system of equations
$$
\left\{\begin{array}{l}
y=4 x^{3}-3 x, \\
z=4 y^{3}-3 y, \\
x=4 z^{3}-3 z
\end{array}\right.
$$

Analysis: Each equation in the system has the same structure as the triple angle formula $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$, so a trigonometric substitution can be used, but $|x|$, $|y|$, and $|z|$ must all be no greater than 1.
Solution: If $|x|>1$, then
$$
y=4 x^{3}-3 x=x\left(4 x^{2}-3\right),
$$

we get $|y|>|x|$.
Similarly, $|z|>|y|,|x|>|z|$, which is contradictory.
Therefore, $|x| \leqslant 1$.
Similarly, $|y| \leqslant 1,|z| \leqslant 1$.
Let $x=\cos \theta, \theta \in[0, \pi]$, then the original system of equations becomes
$$
\left\{\begin{array}{l}
y=\cos 3 \theta, \\
z=\cos 9 \theta, \\
x=\cos 27 \theta .
\end{array}\right.
$$

From $\cos 27 \theta=\cos \theta$, we get
$$
\sin 13 \theta \cdot \sin 14 \theta=0 \text {. }
$$

Thus, $\sin 13 \theta=0$ or $\sin 14 \theta=0$.
Therefore, $\theta=\frac{k \pi}{13}, k=0,1, \cdots, 13$ or $\theta=\frac{k \pi}{14}$, $k=0,1, \cdots, 14$.
Hence, $(x, y, z)=(\cos \theta, \cos 3 \theta, \cos 9 \theta)$.
There are 27 solutions in total.