$3^{n+1}$ divides $2^{3^n}+1$ Describe all positive integers,n such that $3^{n+1}$divides $2^{3^n}+1$.
I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,$3^{n+1}$ divides $2^{3^n}+1$. Kindly clarify this doubt and if it's the former part, please verify my solution-n=1.

We can easily verify that $n=1$ is a solution.
Assume for some $k$ that,
$3^{k+1}|2^{3^{k}}+1$, hence there exists an integer $m$, such that,
$m(3^{k+1})=2^{3^{k}}+1$.
Then we have, 
$2^{3^{k+1}}+1=(2^{3^{k}})^{3}+1=(2^{3^{k}}+1)[(2^{3^{k}})^{2}-2^{3^{k}}+1]=m(3^{k+1})[(2^{3^{k}})^{2}-2^{3^{k}}+1]$. (*)
Now, $3^{k}$ is clearly odd for all integers $k$, hence,
$(2^{3^{k}})^{2}-2^{3^{k}}+1 \equiv ((-1)^{3^{k}})^{2}-(-1)^{3^{k}}+1 \equiv 1+1+1 \equiv 0\mod 3$.
Thus let,
$(2^{3^{k}})^{2}-2^{3^{k}}+1=3n$ where $n$ is an integer.
Continuing from (*), we are left with,
$m(3^{k+1})[(2^{3^{k}})^{2}-2^{3^{k}}+1]=m(3^{k+1})[3n]=m(3^{k+2})(n)$
And so, 
$3^{k+2}|2^{3^{k+1}}+1$.
Thus our inductive proof is complete.
