When is 1! + 2! + … + n! a perfect square?

Small cases. $n=1$: $1=1^2$. $n=2$: $3$, not a square. $n=3$: $1+2+6=9=3^2$. $n=4$: $33$, not a square.

Large $n$. For $k\ge5$, $5\mid k!$. So for $n\ge4$,
$$1!+\dots+n!\equiv1+2+6+24=33\equiv3\pmod5.$$
But squares modulo $5$ are only $0,1,4$, so $3$ is not a square modulo $5$.

Answer. $n=1$ and $n=3$.

Lesson. To show that an equation has no solutions, find a modulus in which the sum stabilises and the target is not a square (here $5$, where the factorials die out).