1. Find all positive integers $x, y$, satisfying
$$
1!+2!+3!+\cdots+x!=y^{2} \text {. }
$$
1. When $x \geqslant 5$, $1!+2!+\cdots+x!\equiv 3(\bmod 5)$, but for any $y \in Z^{+}, y^{2} \equiv 0,1$ or $4(\bmod 5)$, so there is no solution when $x \geqslant 5$. When $x<5$, direct calculation yields two solutions: $x=1, y=1$ and $x=3, y=3$