Use induction

$\begin{aligned} 2^{3^n}+1&= (3-1)^{3^n}+1 \\ &= \sum_{k=0}^{\color{#3D99F6}3^n} (-1)^k {3^n \choose k} 3^{3^n-k} +1 \\ &= \sum_{k=0}^{\color{#D61F06}3^n-1} (-1)^k {3^n \choose k} 3^{3^n-k} {\color{#D61F06}-1}+1 \\ &= \sum_{k=0}^{\color{#D61F06}3^n-1} (-1)^k {3^n \choose k} 3^{3^n-k} \end{aligned}$

We are going to apply induction on n. For $n = 0$ the statement is obvious. Assuming that the claim is true for n consider the following:

$2^{3^{n+1}} + 1 = (2^{3^{n}})^3 + 1 = (2^{3^{n}} + 1)(2^{2 \cdot 3^{n}} - 2^{3^{n}} + 1)$

From $2^{3^{n}} + 1 \equiv 0 \pmod{3^{n+1}}$ and $2^{2 \cdot 3^{n}} - 2^{3^{n}} + 1 \equiv 1 - 2 + 1 \equiv 0 \pmod{3}$ (since $2^{even} \equiv 1$ and $2^{odd} \equiv -1 \pmod{3}$ ) we conclude that $2^{3^{n+1}} + 1 \equiv 0 \pmod{3^{n+2}}$ . Hence our induction is complete.

For simplicity, we'll set $2^{3^{n}}+1=a_n$

Start of Induction: When $n=0$ : $2^1+1$ is divisible by $3^1$

Inductive step: Since we have proven the statement for one $n$ , we have to prove it for $n+1$ , $a_{n+1}$ should be divisible by one more factor of 3 $a_{n+1}=2^{3^{n+1}}+1=2^{3\cdot3^{n}}+1=(2^{3^{n}})^3+1=(2^{3^{n}})^3+1=(2^{3^{n}}+1-1)^3+1=(a_n-1)^3+1=a_n^3-3a_n^2+3a_n-1+1=a_n^3-3a_n^2+3a_n$ Since every part has at least one more factor of 3 than $a_n$ the whole sum has also one more factor of 3, making it divisible by $3^{n+2}$ $\Box$