I considered a similar problem 12 years ago. The eqn,
$$7x^2 + y^2 = 2^n\tag{1}$$
for the special case $x=1$ is called the Ramanujan-Nagell equation. For general $x$, Euler gave a solution in terms of trigonometric functions (which nonetheless yield integer $x,y$). Even more generally, for integer $b>0$,
$$(2^{b+2}-1)x^2 + y^2 = 2^{bm+2}\tag{2}$$
I noticed that its odd solutions $x,y$ are given by, for integer $m>0$,
$$\begin{aligned}
x\, &= \frac{2^{(bm+2)/2}}{h}\, |\sin\big(m\cdot\tan^{-1}(h)\big)|\\
y\, &= 2^{(bm+2)/2}\, |\cos\big(m\cdot\tan^{-1}(h)\big)|\\
h\, &= \sqrt{2^{b+2} - 1}
\end{aligned}\tag{3}$$
where $|n|$ is the absolute value and $\tan^{-1}$ is the arctan function. For example, let $b = 1$, then solutions to,
$$7x^2 + y^2 = \color{blue}{2^{m+2}}$$
$$x_m = 1, 1, 1, 3, 1, 5, 7,\dots$$
$$y_m = 1, 3, 5, 1, 11, 9, 13,\dots$$
which are A077020 and A077021, respectively. Let $b = 2$, then solutions to,
$$15x^2 + y^2 = \color{blue}{2^{2m+2}}$$
$$x_m = 1, 1, 3, 7, 5, 33, 13,\dots$$
$$y_m = 1, 7, 11, 17, 61, 7, 251,\dots$$
where the $x_m$ is A106853 though it is defined there as "the expansion of $\frac{1}{(1-x(1-4x)}$". For $b=3$, the $x_m$ is A145978 and "the expansion of $\frac{1}{(1-x(1-8x)}$". And so on.
Notice that $(2)$ involves the Mersenne numbers $M_n = 2^n-1$. I cannot prove that $(3)$ gives the unique odd solution $x,y$, but it may be the case when $M_n$ is a Mersenne prime.
(P.S. The proven uniqueness for $p=7$ is more due to $\mathbb{Q}(\sqrt{-7})$ being a unique factorization domain, so the uniqueness of odd solutions may or may not extend to other Mersenne primes.)
