Pick an integer $k>1$, and let $n=k^2-1$. We will show that at least one of $n!+1$ and $(n+1)!+1$ can’t be a square.
Indeed, if $n!+1=m^2$ then
$(n+1)!+1 = (n+1)n!+1 = k^2n!+1 = k^2(m^2-1)+1 = (km)^2-n$
Now, clearly $n!+1>n+1$, so $m>k$ and $mk>k^2$. It follows that
$(km-1)^2 = (km)^2-2km+1 < (km)^2-n$
So, finally, the number $(n+1)!+1 = (km)^2-n$ lies strictly between the consecutive perfect squares $(km-1)^2$ and $(km)^2$, so it cannot itself be a perfect square.