find the largest integer $m$ such that $2^m$ divides $3^{2n+2}-8n-9$ 
find the largest integer $m$ such that $2^m$ divides $\space 3^{2n+2}-8n-9$ when $n$ is a natural number.

If the answer was known it will be easy induction.
I started out like this :
$\space 3^{2n+2}-8n-9=9(3^{2n}-1)-8n=9\underbrace{(3^n-1)(3^n+1)}-8n$
Now we have $\frac{3^n-1}{3-1}$ is some integer (sum of GP),or
$ 2|\space 3^n-1$
also we have $3^n+1$ is even ,or
$2|3^n+1....(3)$
From this we conclude $4|(3^n-1)(3^n+1) ...(1)$
Let n be even then $3^n-1=3^{2m}-1=(3^m-1)(3^m+1)$,
by $(1)$ :
$4|(3^m+1)(3^m-1)$ meaning $4|3^n-1...........(2)$
combining $(2),(3)$ we have $8|3^{2n+2}-8n-9$
Similarly i was able to work out the same when $n=2m+1$ by noting that $3^n+1=3^{2m+1}+1$ is divisible by $4$.
I got the largest integer as $3$.
But i am wrong as the MCQ did not have the option $m=3$
how do i proceed.
Note: I have  not learnt about fermat's little theorem
Also i am looking for Hints  rather than complete solutions .use of >!  may help

Hint: For $n=1$, it is clear which is the largest power of $2$.  Now consider $(8+1)^{n+1} - [(n+1)\cdot8+1]$ and use binomial expansion to conclude it works for all larger $n$.
