$12 \cdot 139$ Find the natural number solutions of the equation $1!+2!+\cdots+(x+1)!=y^{z+1}$.
(Former Yugoslavia Mathematical Olympiad, 1975)
[Solution] Let $f(x)=1!+2!+\cdots+(x+1)!$, then
$$
f(1)=3, f(2)=9=3^{1+1}, f(3)=33=3 \cdot 11.
$$

When $x>3$,
$$
f(x)=f(3)+5!+\cdots+(x+1)!=3(\bmod 5).
$$

For any integer $k$,
$$
\begin{array}{l}
(5 k)^{2}=25 k^{2} \equiv 0 \quad(\bmod 5), \\
(5 k \pm 1)^{2}=25 k^{2} \pm 10 k+1 \equiv 1 \quad(\bmod 5), \\
(5 k \pm 2)^{2}=25 k^{2} \pm 20 k+4 \equiv 4 \quad(\bmod 5).
\end{array}
$$

Therefore, $f(x)$ is not a perfect square.
Thus, when $z=1$, for any natural number $x \neq 2$ and $y$, it does not satisfy $f(x)=y^{2}=y^{1+1}$.
When $z \geqslant 2$,
$$
\begin{array}{c}
f(1)=3, f(2)=9, f(3)=33, f(4)=153, \\
f(5)=873, f(7)=46233.
\end{array}
$$

Therefore, when $x=1,2,3,4,5,7$, $f(x)$ is divisible by 3 but not by 27, hence
$$
f(x) \neq y^{z+2}, z \geqslant 2.
$$

When $x>7$,
$$
f(x)=f(7)+9!+\cdots+(x+1)!\equiv f(7) \quad(\bmod 27).
$$

When $x=6$,
$$
f(6)=5913=3^{4} \cdot 73.
$$

Thus, when $x=6$ and $x>7$, $f(x)$ cannot be expressed in the form $y^{2+1}(z \geqslant 2)$. That is, when $z \geqslant 2$, the equation $f(x)=y^{z+1}$ has no natural number solutions.
From the above, the equation has a unique natural number solution
$$
x=2, y=3, z=1.
$$