I was solving problem 137 of Project Euler, which led me to find $n$ such that $5n^2+2n+1$ is a perfect square. But such numbers are very rare (the 13th is around 3 billions) so after decomposing into $(n+1)^2 + (2n)^2 = m^2$ and looking for Pythagorean triples and their $(a^2-b^2,2ab,a^2+b^2)$ generation, I ended up having to look for $k$ such that $5k^2+4$ is a perfect square.
This is a much easier task, which retrospectively makes sense since every $k$ will lead to some $n=O(k^2)$, so you only need to iterate to $10^5$ to find the $13$th number.
So we proved that there are more squares in $5n^2+4$ than in $5n^2+2n+1$.
Was there an easier way to spot this without pulling out the Pythagorean triple trick? Is there an intuitive reason or more generic underlying principle solely by looking at the equations?