$3^{n+1}$ 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.

This actually looks like an induction proof. Let $P(n)$ be the statement : $$3^{n+1} \ \mid \ 2^{3^n} + 1 $$
Then we can verify that it is indeed true for $n = 1$ since $9$ divides itself. So suppoose $P(n)$ is true. Now, 
$$2^{3^{n+1}} + 1 = \left({2^{3^n}}\right)^3 + 1 $$
Notice that $$\left({2^{3^n} + 1}\right)^3 = \left({2^{3^n}}\right)^3 + 3 \cdot \left({2^{3^n}}\right)^2 + 3 \cdot \left({2^{3^n}}\right) + 1 $$
So, 
\begin{align}
\left({2^{3^n}}\right)^3 + 1 & =  \left({2^{3^n} + 1}\right)^3  -  3 \cdot \left({2^{3^n}}\right)^2 - 3 \cdot \left({2^{3^n}}\right) \\
& =  \left({2^{3^n} + 1}\right)^3 - 3 \cdot \left[{\left({2^{3^n } + 1}\right)^2 - 2\left({2^{3^n}}\right) - 1}\right] - 3 \cdot \left({2^{3^n} }\right) \\
& = \left({2^{3^n} + 1}\right)^3  - 3 \cdot \left({2^{3^n } + 1}\right)^2 + 3 \cdot \left({2^{3^n } + 1}\right) \\ 
& = \left({3^{n+1} \cdot k}\right)^3 - 3 \left({3^{n+1} \cdot k}\right)^2 + 3 \left({3^{n+1} \cdot k}\right) \\
& = 3^{n +2} \left({3^{2n+1} \cdot k^3 - k^2 + k}\right)  
\end{align}
which implies $$3^{n+2} \ \mid \ 2^{3^{n+1}} + 1 $$
And so the result is true for every $n$. 
So all the integers satisfy the given condition and that is the answer to the relevant question. 
