I’m trying to solve the Diophantine equation \begin{align} \tag{$\star$} (2^k-3)b^2 - (2^{k+1}+1)ab - (2^k+2)a^2 = 0, \end{align} where $k$ is a positive integer, and $a$ and $b$ are relatively prime integers (not necessarily positive). A computer search turns up only the solutions $k=1$ and $k=4$, with no more solutions $1 \le k \le 100$.
I had hoped that I could use Vieta-jumping to solve it, but after much trying and many sheets of sketch paper, I haven’t found the magic incantation. Any help — either with a Vieta-jumping hint, or with any hints toward a proof mechanism — would be greatly appreciated.
In case it helps, I know that $$ (b^2-2ab-a^2) \mid 2 \cdot 23. $$ Also, it is trivial to show that ($\star$) implies $2^{2k+3}=c^2+23$ for some integer $c$.
You have a minor variant of Ramanujan Nagell. I would imagine a complete resolution is available somewhere.
https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Nagell_equation