Infinitely many positive integers $n$ such that $n^2+1 \mid 1 \cdot 2 \cdot 5 \cdot 10 \cdots ((n-1)^2+1)$ 
Prove that there exist infinitely many positive integers $n$ such that $$n^2+1 \mid 1 \cdot 2 \cdot 5 \cdot 10 \cdots ((n-1)^2+1).$$

We need to prove that there are infinitely many $n$ such that $\prod_{m=0}^{n-1}(m^2+1)$ is divisible by $n^2+1$. Some examples I found for when it is divisible are $n = 3,7,8,13,17,18,21$, but I didn't see how to find infinitely many.

I conjecture that $n=2^{2k+1}$ is always a solution. We have that $$2^{4k+2}+1=(2^{2k+1}+1)^2-2^{2k+2}=(2^{2k+1}-2^{k+1}+1)(2^{2k+1}+2^{k+1}+1)$$
Now we can pick $m=2^{k+1}+1$ then $$m^2+1=2^{2k+2}+2^{k+2}+2=2(2^{2k+1}+2^{k+1}+1)$$ and $m=2^{k+1}-1$ then $$m^2+1=2^{2k+2}-2^{k+2}+2=2(2^{2k+1}-2^{k+1}+1)$$
And both $m$'s are obviously part of the RHS product (if $k>1$) the conjecture is true hence infinitely many solutions.
