Prove $2^{n+3}|(p^2-1)$ Olympiad problem:

let odd prime nunmber $p$ such $$p|(2+\sqrt{3})^{2^n}+(2-\sqrt{3})^{2^n}$$
  show that:
  $$2^{n+3}|(p^2-1)$$

I want to take advantage of this conclusion：
if prime factor $ p $ of a Fermat number $ 2^{2^n} + 1 $ ,then  $ p \equiv 1 \pmod{2^{n + 2}} $ 
But I failed

If $p\equiv 1\pmod{2^{n+2}}$, then either $p\equiv 1\pmod{2^{n+3}}$ or $p\equiv 2^{n+2}+1\pmod{2^{n+3}}$. In either case, squaring yields $p^2\equiv 1\pmod{2^{n+3}}$.
