We will construct an infinity of numbers in which the requirement of the problem is verified.
Consider n+1−a=p2.So n2+n+a=(n+1)2−n−1+a=(p2+p+a)⋅(p2−p+a).
It is clear that for sufficiently large p we have a<p2+p+aa<p2+p+a2<p2−p+a .
If a>1 we take p\equiv-1 \pmod a sufficiently large so p^2+p+a\equiv 0 \pmod a and 1<a<\frac{p^2+p+a}{a}<p^2-p+a<p^2+a-1=n so
(p^2+a-1)! is divisible by (p^2+p+a)\cdot(p^2-p+a) for infinitely many p.So n! is divisible by n^2 + n + a for infinitely many positive integers n..
If a=1 we have that n=p^2.We take p \equiv 1 \pmod 3 sufficiently large so 1<3<\frac{p^2+p+1}{3}<p^2-p+1<p^2=n so (p^2)! is divisible by (p^2+p+1)\cdot(p^2-p+1) for infinitely many p.So n! is divisible by n^2 + n + 1 for infinitely many positive integers n.