The present question is directly inspired by this one.
Let $\alpha$ be a unit in the ring of quadratic integers of a real quadratic field, or, in less sophisticated words: $$\alpha=\frac{a\pm\sqrt{a^2\pm4}}{2}, $$ for some natural number $a$.
Let $\bar\alpha$ be the conjugate of $\alpha$, that is: $\overline{\frac{a\pm\sqrt{a^2\pm4}}{2}}=\frac{a\mp\sqrt{a^2\pm4}}{2} $.
Now, for $n$ a natural number, let $F_n(\alpha)$ be a sort of generalized Fermat number, such that $$ F_n(\alpha):= \alpha^{2^n}+\bar\alpha^{2^n}. $$
$F_n(\alpha)$ is a natural number, since in the expansion of $F_n(\alpha)$ all the square roots cancel each other.
Let $p$ be an odd prime divisor of $F_n(\alpha)$. Is it true that $$ p^2 \equiv 1 \pmod {2^{n+1} } $$ Edit 21/12/18: More pecisely, is it true that
$$ p\equiv \left(\frac{a^2\pm 4}{p}\right) \pmod {2^{n+1} } $$ where $ \left(\frac{\cdot}{p}\right)$ is the Legendre symbol.
As an example, with $\alpha = 2+\sqrt5$, we have:
$$ F_4(\alpha)= 10749957122=2\cdot769\cdot2207\cdot3167 $$ and \begin{align*} 769 &\equiv \left(\frac{5}{769}\right) = +1\pmod {2^5}\\ 2207 &\equiv \left(\frac{5}{2207}\right)=-1 \pmod {2^5}\\ 3167 &\equiv \left(\frac{5}{3167}\right)=-1 \pmod {2^5} \end{align*}
I have checked many other $\alpha$ and $n$, but I could not find a counterexample to this.
Edit 21/12/18: a proof should be like in the answer to the above related question.