Example 6 Find all positive integers $x, y'$, satisfying
$$
1!+2!+3!+\cdots+x!=y^{2} \text {. }
$$
Solution: Since when $x \geqslant 5$, $5!+\cdots+x!=0$ $(\bmod 5)$, and $1!+2!+3!+4!=33=3$ $(\bmod 5)$, hence at this time $1!+2!+3!+\cdots+x$ ! $=3(\bmod 5)$. However, for any $y \in Z^{1}, y^{2} \equiv 0$ or $1$ or $4(\bmod 5)$. Therefore, when $x \geqslant 5$, the equation has no positive integer solutions.
If $x<5$, by direct calculation we get the only solutions:
$x=1, y=1$ and $x=3, y=3$.