4. All positive integer pairs $(x, y)$ that satisfy $1!+2!+\cdots+x!=y^{2}$ are $\qquad$
$$
\text { 4. }(1,1),(3,3) \text {. }
$$

Since when $x \geqslant 5$, we have
$$
\begin{array}{l}
1!+2!+\cdots+x! \equiv 3(\bmod 5), \\
y^{2} \equiv 0,1,4(\bmod 5),
\end{array}
$$

thus, $x<5$.
By direct calculation, we know $(x, y)=(1,1),(3,3)$.