Example 5 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 $3 \mid y$, and thus 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 $n \geqslant 9$, $3^{3} \mid n!$. Therefore, when $x \geqslant 8$, the left side of (1) is divisible by $3^{2}$ but not by $3^{3}$, and 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. Modulo 5: when $x \geqslant 8$, the left side of (1) $\equiv$ $1!+2!+3!+4!\equiv 3(\bmod 5)$; and since it has been proven that $z=2$ in this case, the right side of (2) $z^{2} \equiv 0$, or $1(\bmod 5)$, thus the above assertion holds.

Finally, when $x<8$, it is not difficult to find through inspection that the solution to (1) is $x=y=3, z=2$.