$3^{n+1}$ 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.

By induction: case $n=0,1$ is obvious, assume the claim for $n \in \mathbb{N}$. Then, $$2^{3^{n+1}} +1 = ((2^{3^{n}}+1)-1)^3 +1 = (2^{3^{n}}+1)^3 -3(2^{3^{n}}+1)^2 +3 (2^{3^{n}}+1)$$ and by the induction hypothesis $ 3^{n+2}$ divides the last two terms. For the first term (call it $a$), induction again gives $3^{n+1}$ divides $a^{1/3}$. Then, $3^{n+2}$ divides $3^{3n+3}$ which divides $a$ so by transitivity we're done.

It follows from one of the Lifting The Exponent Lemmas (LTE):

Let $\upsilon_p(a)$ denote the exponent of the largest power of $p$ that divides $a$.

If $p$ odd prime, $n\in\mathbb Z^+$ is odd, $a,b\in\mathbb Z$, $a\equiv -b\not\equiv 0\pmod{p}$, then

$$\upsilon_p\left(a^n+b^n\right)=\upsilon_p(a+b)+\upsilon_p(n)$$

Therefore $\upsilon_3\left(2^{3^k}+1\right)=\upsilon_3(2+1)+\upsilon_3\left(3^k\right)=k+1$. So in fact, I've proved a stronger result: We have $3^{k+1}\mid 2^{3^k}+1$ and $3^{k+2}\nmid 2^{3^k}+1$, for all $k\in\mathbb Z_{\ge 0}$.