The many, many, many stars and a pinch of stripes for good measure

In the "Huh, that's neat ... I guess" category of number theory, the preamble to this week's Riddler Classic notes that the current number of states satisfies that $N$ is double a square and that $N+1$ is a centered pentagonal number.

After $50,$ what is the next integer $N$ with these properties?

After looking up the definition of a centered pentagonal number, it appears that they are of the form $$1 + \sum_{k=1}^n 5k = 1 + 5 \frac{n(n+1)}{2}.$$ So the question states find the minimal $N$ such that there exists integers $n$ and $m$ with $$N = 2n^2 \,\, \text{and} \,\, N+1 = 5\frac{m(m+1)}{2} + 1.$$ In particular, then we would have $$m^2 + m - \frac{4}{5} n^2 = 0$$ for some integers $m, n \in \mathbb{N}.$ Solving for $m$ in terms of $n$ we get $$m = \frac{-1 \pm \sqrt{1 + \frac{16}{5} n^2 }}{2}.$$ In order for $m$ to be an integer we would need $\sqrt{1 + \frac{16}{5}n^2}$ to be an odd integer, so in particular, we require that $5 \mid n.$

So we can brute force check whether the conditions hold for each multiple of $5$, until arriving at $n = 90,$ which gives $\sqrt{1 + \frac{16}{5} n^2} = 161$ and a grand total of $N = 2(90)^2 = 16,200$. This seems like either we would need much bigger flags, or much smaller states to accommodate a flag with those specifications. And at what point on our walk from $51$-ish states to $16,200$ would we reconsider the convention of only having 13 stripes for the original colonies?