I conjecture that there exist infinitely many integers $n$ such
that $$(n^{2015}+1)\mid n!.$$
I have seen a simpler problem that there exist infinitely many integers $n$ such that $(n^2+1)\mid n!$.
Alternatively, I considered the Pell equation
$n^2+1=5m^2$, $2m<n$, but for $2015$ I can't figure it out.
Best Answer
Modest progress. There are infinitely many integers $n$ such that $n^3+1\mid n!$.
We always have $n^3+1=(n+1)(n^2-n+1)$. Let $n=k^2+1$. Then $$ n^2-n+1=(1+k+k^2)(1-k+k^2). $$ Assume further that $k\equiv1\pmod3$. In that case $1+k+k^2$ and $n+1=2+k^2$ are both divisible by $3$. For all sufficiently large $k\equiv1\pmod3$ we thus have $$ (k^2+1)^3+1=3^2\cdot\frac{k^2+2}3\cdot\frac{k^2+k+1}3(k^2-k+1) $$ that is clearly a factor of $(k^2+1)!$.