Given fundamental discriminant $d \equiv -1 \bmod 8$ such that the quadratic imaginary number field $\mathbb{Q}(\sqrt{-d})$ has odd class number $h(-d)$. Is it true that one can always solve the diophantine equation,
$$x^2+dy^2 = 2^{2+h(-d)}\tag{1}$$
with odd integers $x,y$?
For example, one can indeed solve,
$$x^2+7y^2 = 2^3$$
$$x^2+23y^2 = 2^5$$
$$x^2+47y^2 = 2^7$$
$$x^2+71y^2 = 2^9$$
which have $d$ with $h(-d) = 1,3,5,7$, respectively, and it is solvable for all such $d$ with $h(-d) \leq 25$ in the link given above (I tested them all), but it would be nice to know if it is true in general.