Example 2 Find all positive integers $x>1, y>1, z>1$, such that $1!+2!+\cdots+x!=y^{z}$.
The key step is to prove that when $x \geqslant 8$, it must be that $z=2$. Since the left side of (1) is divisible by 3, hence $3 \mid y^{z}$, which implies $3 \mid y$, so the right side of (1) is divisible by $3^{z}$. On the other hand,
$$1!+2!+\cdots+8!=46233$$

is divisible by $3^{2}$, but not by $3^{3}$; and for $n \geqslant 9$, $3^{3} \mid n!$, so when $x \geqslant 8$, the left side of (1) is divisible by $3^{2}$, but not by $3^{3}$, thus the right side of (1) must also be so, i.e., $z=2$ must hold.

Now, we further prove that when $x \geqslant 8$, equation (1) has no solutions. Taking both sides of (1) modulo 5, when $x \geqslant 8$, the left side of (1) $\equiv 1! + 2! + 3! + 4! \equiv 3 (\bmod 5)$; and since $z=2$ is known, the right side of (1) $\equiv 0, \pm 1 (\bmod 5)$, thus the above assertion holds. Finally, when $x<8$, by verification, the solution to (1) is $x=y=3, z=2$.