Prove that if $n$ is a positive integer, then $3^{2^{n}}-1$ is divisible by $2^{n+2}$
I tried to prove this using induction so for the base case $n=1$ you get that $8 \mid 8$
For the inductive step assume $P(k): 2^{k+2}\mid 3^{2^{k}}-1$ is true for some $k \geq 1$ I want to show $P(k+1)$ is true then $2^{k+2+1}=( 3^{2^{k}}-1) * 2$ I want to somehow get to $3^{2^{k+1}}-1$ I just got stuck and I'm unsure if this is the right approach.
Hint:$$3^{2^{k+1}}-1 = \left(3^{2^{k}}\right)^2-1=(3^{2^{k}}-1)(3^{2^{k}}-1+2).$$
For some integer $k$: $\ 3^{2^n}-1=2^{n+2}k$ and from this you want to derive, for some integer $m$: $\ 3^{2^{n+1}}-1=2^{n+3}m$.
$$3^{2^n}3^{2^n}-1=2^{n+3}m$$ $$(2^{n+2}k+1)(2^{n+2}k+1)-1=2^{n+3}m\Longrightarrow$$ $$2^{2n+4}k^2+2^{n+3}k+1-1=2^{n+3}m\Longrightarrow$$ $$2^{n+1}k^2+k=m$$
Because $k$ is an integer $m$ is an integer.
As a test, if $n=1$ then $k=\frac88=1$, so when $n=2$ then $m=2^2\cdot 1+1=5=\frac{80}{16}$
