Prove that there exist infinitely many integers $(n^{2015}+1)\mid n!$ 
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.

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 $(k^2+1)!$.
