The sequence $x_0,x_1,x_2,x_3,x_4,…..$ is defined as $x_0=2$,$x_{k+1}=2cdot x_k^2-1$,for every $kgeq0$.Prove that if an odd prime number $p$ divides $x_n$,then $2^{n+3}$ divides $p^2-1$

My idea was to firstly treat simple cases like this:

If $n=0$,then $x_0=2$ doesn’t have any odd divisors.

If $n=1$,then $x_1=7$.It has only one odd prime divisor,$7$ and $2^{n+3}=2^4=16$ divides $p^2-1=7^2-1=48$

………

From here, my idea was to aply some sort of induction but induction doesn’t work too well with prime numbers so this is where I got stuck with this idea.

Another idea was to write $x_n$ in its polynomial form and making it easier to prove what we want.I think that induction might work here but I stil didn’t mange to do it.Can you help me ,please?