Solve $\because 1!=1,1!+2!=3$, $1!+2!+3!=9, \quad 1!+2!+3!+4!=33$.
When $n \geqslant 5$, the last digit of $n!$ is 0,
$\therefore$ the last digit of $1!+2!+3!+\cdots+n!$ is 3.
According to the property of square numbers (1). It must be $n<5$, and it is evident that the integer
solutions of the equation are $\left\{\begin{array}{l}x=1,3, \\ y=1,3,\end{array}\right.$