Prove that there are infinitely many integers n such that $4n^2+1$ is divisible by both $13$ and $5$

$$4n^2 + 1 \equiv 0 \pmod{13} \iff n^2 \equiv 1 \pmod 5 \iff n \equiv \pm 1 \pmod 5$$

and

$$4n^2 + 1 \equiv 0 \pmod{13} \iff n^2 \equiv 3 \pmod{13} \iff n \equiv \pm 4 \pmod{13}.$$

In conclusion, $$4n^2 + 1 \equiv 0 \pmod{65} \iff n \pmod{65} \in \{4, 9, 51, 56\}.$$

The solutions are $$\{4 + 36k, 9 + 36k, 51 + 36k, 56 + 36k | k \in \mathbb{Z}\}.$$

I'm surprised no one has mentioned the OEIS, since this was available at the time this question was asked:

A203464 Numbers $n$ such that $65$ divides $4n^2 + 1$; alternately, numbers which are $4, 9, 56$, or $61 \bmod 65$.

$4, 9, 56, 61, 69, 74, 121, 126, 134, 139, 186, \ldots$

The sequence is infinite, since every number of the form $65k + 56$ is a member. - Arkadiusz Wesolowski, Oct 29 2013

And there isn't the "fini" keyword anywhere to be seen.

Don't just take Mr. Wesolowski's word for it, though. Verify that $(65k + 56)^2 = 4225k^2 + 7280k + 3136$. Doubling twice, we obtain $16900k^2 + 29120k + 12544$. Clearly $16900k^2$ and $29120k$ are both multiple of $65$. But $12544$ is not. In fact, it's $1$ short of a multiple of $65$. Q.E.D.

Of course this doesn't account for all the numbers listed. But it is sufficient to prove the sequence is infinite.

$n = 13 \times 10^k + 4$ implies $4n^2 + 1 = 4(13^2 \times 10^{2k} + 8 \times 13 \times 10^k + 16) + 1$, which in turn is $4 \times 13^2 \times 10^{2k} + 32 \times 13 \times 10^k + 65$. It is clear that all terms are multiples of $13 \times 5 = 65$.