A problem about divisibility

Show that there exists infinitely many positive integers $\displaystyle n $ such that $\displaystyle n^2+1 | n! $ .

I am begging for seeing all $\displaystyle n < 1000 $ satisfying the above requirement . The job should belong to computer but i have no idea how to design a computer programme to look for the numbers .

Here's my attempt to the proof :

Let $\displaystyle a_n $ be the sequence with $\displaystyle a_1 = 21 ~,~ a_{n+1} = a_n^2 - a_n + 1 $ . Then i find that $\displaystyle a_n \equiv 1 \bmod{5} ~,~ n\geq 1 $ . Moreover , interestingly i can factorize $\displaystyle a_{n+1}^2 + 1 $ into $\displaystyle ( a_n^2 + 1 ) ( (a_n - 1 )^2 + 1 ) $ . But $\displaystyle (a_n-1)^2 + 1 < a_{n+1} $ and $\displaystyle (a_n-1)^2 + 1 $ is prime to $\displaystyle a_n^2 + 1 $ . Therefore , if $\displaystyle a_n^2 + 1 | (a_n)! $ , we also have $\displaystyle a_{n+1}^2 + 1 | (a_{n+1})! $ then it finishes the proof by induction .

Since the sequence contains the 'very large' integers ( $\displaystyle a_4 = 31265489221$ !) , i wonder if it is true that the numbers could hardly be found in the integer line .... But i think that is not true , i believe by solving quadratic congruences we can construct many such integers !

Thank you