It is still unknown if $n!+1=m^2$ has only 3 solutions, i.e. for $n=4,5,7$.
https://en.wikipedia.org/wiki/Brocard%27s_problem
What about my problem:
Are there finitely or infinitely many numbers $n$ such that $n!+1$ is divisible by a perfect square greater than 1?
It is also interesting if $n!+1$ is squarefree for infinitely many $n$.
For $n\le 100$, number $n!+1$ is:
prime for $n=1,2,3,11,27,37,41,73,77$
perfect square for $n=4,5,7$
divisible by a square (while not being a perfect square) for $n=12,23$
composite squarefree for other $n$'s