How to prove the existence of infinitely many $n$ in $\mathbb{N}$,such that $(n^2+k)|n!$ 
Show there exist  infinitely many $n$ $\in \mathbb{N}$,such  that
  $(n^2+k)|n!$ and  $k\in N$

I have a similar problem:

Show that there are infinitely many $n \in \mathbb{N}$,such  that
  $$(n^2+1)|n!$$

Solution: We consider this pell equation,$n^2+1=5y^2$,and this pell equation have
$(n,y)=(2,1)$,so this equation have infinite solution$(n,y)$,and 
$2y=2\sqrt{\dfrac{n^2+1}{5}}<n$.
so $5,y,2y\in \{1,2,3,\cdots,n\}$,
so $5y^2<n!$
then $(n^2+1)|n!$
 but for $k$ I have consider pell equation,But I failed,Thank you everyone can help

Similar to your solution of $k=1$.
Consider the pell's equation $n^2 + k = (k^2+k) y^2$. This has solution $(n,y) = (k,1)$, hence has infinitely many solutions. Note that $k^2 + k = k(k+1) $ is never a square for $k\geq 2$, hence is a Pell's Equation of the form $n^2 - (k^2+k) y^2 = -k$.
Then, $2y = 2\sqrt{ \frac{ n^2+k} { k^2 +k } } \leq n$ (for $k \geq 2$, $n\geq 2$) always.
