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.