$ 3^{2^n }- 1 $ is divisible by $ 2^{n+2} $ Prove that if n is a positive integer, then $ \ \large 3^{2^n }- 1  $ is divisible by  $ \ \large 2^{n+2} $ .
Answer:
For  $ n=1 \ $ we have 
$ \large 3^{2^1}-1=9-1=8 \ \ an d \ \ 2^{1+2}=8 $
So the statement hold for n=1. 
For $ n=2 $ we have 
$ \large 3^{2^2}-1=81-1=80  \ \ and \ \ 2^{2+2}=16 \ $
$Also \ \ \  16 /80 $ . 
Thus the statement hold for $ n=2 $ also. 
Let the statement hold for $ n=m \ $
Then $ a_m=3^{2^m}-1 \ \ is \ \ divisible \ \ by \ \ b_m=2^{m+2} \ $ 
We have to show that $ b_{m+1}=2^{m+3}  \ $ divide $ \ \large a_{m+1}=3^{2^{m+1}}-1 \ $ 
But right here I am unable to solve . If there any help doing this ? 
Else any other method is applicable also.

$$\begin{array}{} &3^{2^m} -1 &=x 2^{m+2} & \text{assumed and checked by } m=1 \text{with $x$ odd}\\ 
 &3^{2^{m+1}} -1 &= 3^{2 \cdot 2^m} -1  &  \text{general step in induction} \\ 
 & &= (3^{2^m})^2 -1\\
 & &= (3^{2^m} -1)(3^{2^m} +1) \\
 & &=x 2^{m+2}(3^{2^m} +1) & \text{ with $x$ odd }\\
 & &=x 2^{m+2}(3 \cdot 3^{2^m-1} +1) \\
 & &=x 2^{m+2}(2 \cdot 3^{2^m-1}+ (3^{2^m-1}+1)) \\
 & &=x 2^{m+2}(2 \cdot 3^{2^m-1}+ (3+1) \cdot y) & \text{by $3^{2w+1}+1$}=(3+1)\cdot y\\
 & &=x 2^{m+2}2 \cdot (3^{2^m-1}+ 2 y) &\text{by factoring out }2\\
 & &=x 2^{m+3} z &\text{with $x$ and $z$ odd}\\
\end{array}$$
$$\begin{array}{l}\implies  2^{m+3} \mid 3^{2^{m+1}}-1 & \phantom {\text{yxcyxcyyxcyxcycyxcyc}}& \text{induction successful, proof complete}
\end{array}$$
