I am investigating when a Proth number is a perfect square. I saw a very short proof in this article (Lemma 3.1), but I want it more number-theoretic.
A Proth number $n$ is a number of the form $$ n = h\cdot 2^k + 1 \text{ with } k\in\mathbb N \text { and } h<2^k \text { odd.} $$ So lets suppose we have $h\cdot 2^k + 1 = q^2$ for a $q\in\mathbb N$. Then $$ h\cdot 2^k = q^2 - 1 = (q+1)\cdot (q-1). $$ Since $h\cdot 2^k$ is even, so is $(q+1)(q-1)$. That means that both $q+1$ and $q-1$ are even. Since their difference is $2$, it follows that one of these is divisible by $2$ exactly once. And the other is divisible by $2$ exactly $2^{m-1}$ times.
But how do I go on from here?