I have a proof of this simple problem, but I feel that the last step is rather clunky:
For $n=1,2,3,4$ we have $n!+5=6,7,11,29$ respectively, none of which are square. Now assume that $n\geq 5$, then: $$\begin{aligned} n! +5 & \;=\; n(n-1)\cdots 6\cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 + 5 \\[0.2cm] & \;=\; 5\left[ \frac{}{} n(n-1)\cdots 6\cdot 4 \cdot 3 \cdot 2 \cdot 1 + 1 \right] \\[0.2cm] &\;=\; 5(3k+1) \end{aligned}$$ for some $k\in\mathbb{N}$. Since $5(3k+1)=15k+5\equiv 5\,\text{mod} \, 15$ and all perfect squares are congruent either $0,1,4,6, 9$ or $10\,\text{mod}\,15$, the result follows. $\;\blacksquare$
It took a pretty tedious exhaustive search of squares modulo 15 in the last step; is there a theorem I am missing that means the last step follows immediately? Your comments are much appreciated!