How prove this there many infinite numbers postive ineteger $n$ such $n^3+1|n!$ Question:

show that: there exsit infinite postive integer $n$ such

$$n^3+1|n!$$
Buy the way,I know prove this  this simple problem:
prove there are many infinte postive  integer numbers $n$,such 
$n^2+1|n!$ 
This problem we only  consider this pell equation
$$n^2+1=5y^2$$ have a solution $(2,1)$, so the pell equation have many infite solution.
because $$2y=2\sqrt{\dfrac{n^2+1}{5}}<n\Longrightarrow n>5, 5,y,2y\in\{1,2,3,\cdots,n\}$$
so there exsit infinte $n$ such
$$n^2+1|n!$$
But this is $n^3+1$ so I can't .Thank you 

I will prove there are infinitely many solutions of the form $n=3a^2$. For such $n$ we have $3 \nmid n+1$, hence $\gcd(n+1,n^2-n+1)=1$ and it is sufficient to show that both $n+1 \mid n!$ and $n^2-n+1 \mid n!$. We have $n+1 \mid n!$ whenever $n+1 = 3a^2+1$ is not prime, whereas $n^2-n+1 = 9a^4-3a^2+1$ factors as $(3a^2+3a+1)(3a^2-3a+1)$. Since both factors are relatively prime, we should have $3a^2+3a+1 \mid (3a^2)!$ and $3a^2-3a+1 \mid (3a^2)!$. The latter is always true, whereas $3a^2+3a+1 \mid (3a^2)!$ is true when $3a^2+3a+1$ is not prime. Therefore, it is sufficient to show that there are infinitely many $a$ for which neither $3a^2+1$ nor $3a^2+3a+1$ is prime, which should not be that hard (e.g. take $a \equiv 1 \mod 7$ and $a \equiv 2 \mod 13$).
