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.
 A: Since
$2015 = 5\cdot 13\cdot 31
$,
and
$n^a + 1| n^{ab}+1
$
if $b$ is odd,
a necessary condition for
$n^{2015}+1 | n!
$
is
$n^m+1 | n!$
for every $m$
in
$\{5, 13, 31
, 5\cdot 13
, 5\cdot 31
, 13\cdot 31
\}
$.
Solutions are going to be hard to find.
All those expressions
of the form
$n^j-n^{j-1}+...-n+1
$
for odd $j$
will have to have
all prime factors
$\le n$
in order to divide
$n!$.
A: 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 $(k^2+1)!$.
A: It suffices to show that for infinitely many $n$, the largest prime factor of $n^{2015}+1$ is at most $\sqrt{n}$. Indeed, if $n$ is such a large integer and $p$ is a prime, then the largest value of $a$ for which $p^a\mid n^{2015}+1$ is $\leq c \log n$ for some constant $c$, while $n!$ is divisible by $p^a$ with $a\geq \frac{n}{p}-1\geq \sqrt{n}-1>c \log n$. It was shown by Schinzel (Theorem 13 in https://eudml.org/doc/204826) that for any nonzero integers $A$ and $B$, any integer $k\geq 2$ and any $\varepsilon>0$ there exist infinitely many integers $n$ such that the largest prime factor of $An^k+B$ is less than $n^{\varepsilon}$. In particular, the claim of the problem holds with $2015$ replaced by any positive integer.
