Solutions of $x^2 + 7y^2 = 2^n$ where $x$ and $y$ are odd numbers Is it true that for any $n\geq 2$ the equation $x^2 + 7y^2 = 2^n$ has a solution with $x$ and $y$ odd ??

The values of $2(\frac 12 +\frac 12 \sqrt{-7})^{n-2}$ give $x+y\sqrt{-7}$, which are a solution to the equation $x^2+7y^2 = 2^n$, where $x$ and $y$ are always odd.  
These solutions are thence unique, because all representations of $x^2+7y^2$ reduce to a complex number $x+y\sqrt{-7}$, and that $\frac 12 \pm \frac 12 \sqrt{-7}$ is the unique primes of this form.  $2$ is the product of these numbers, and since it requires both primes to give an even value of $x$ and $y$, powers of a single prime must lead always to odd numbers.
