$3^{n+2}$ does not divide $2^{3^n}+1$ Prove for all natural numbers $n$ such that $2^{3^n}+1$ is divisible by $3^{n+1}$, but is not divisible by $3^{n+2}$.
I know how to prove why is it divisible, but I need help with why is it not divisible.
I used a similar proof for why is it divisible:

https://math.stackexchange.com/a/1591199/925421

My working for not divisible: Induction proof
Base case: n = 0
$$3^{0+2}=9$$
and
$$2^{3^0}+1=3$$
9 cannot divide 3, so base case is true.
Assume n=k is true. That means $3^{k+2}$ does not divide $2^{3^k}+1$. Now prove $3^{k+3}$ does not divide $2^{3^{k+1}}+1$.
$$(2^{3^k})^3+1=(3^{k+1}*m)^3-3(3^{k+1}*m)^2+3(3^{k+1}*m)$$
$$=3^{3k+3}*m^3-3^{2k+3}*m^2+3^{k+2}*m$$
$$=3^{k+3}(3^{2k}*m^3-3^{k}*m^2+3^{-1}*m)$$

That's where I need help!
I'm so close to finish, I want to show that $$(3^{2k}*m^3-3^{k}*m^2+3^{-1}*m)$$ is not an integer, so that means $3^{n+3}$ does not divide $2^{3^{n+1}}+1$

To show that $(3^{2k}*m^3-3^{k}*m^2+3^{-1}*m)$ is not an integer, I have to show m can never be 3 or multiple of 3. But I don't know how to show that. Please help.

The linked problem also has an answer involving the Lifting The Exponent (LTE) result. Here is a plain solution using only induction for the proposition:
$$
\color{blue}{
P(n)\text{ : We have $2^{3^n}=-1 + 3^{n+1}(1+3N(n))$ for some integer $N(n)$ .}
}
$$
Let us inductively show that $P(n)$ is true for all integers $n\ge 0$.
$P(0)$ is true, $2^{3^0}=2^1=2=-1+3=-1+3^{0+1}$ and we take $N(0)=0$.
$P(1)$ is true, $2^{3^1}=2^3=9=-1+3^2=-1+3^{1+1}$ and we take $N(1)=0$.
Fix some $n\ge 1$. Assume that (for this $n$) the proposition $P(n)$ is true. Then, working modulo
$3^{(n+1)+2}$ at the places with the $\equiv$ relation symbol:
$$
\begin{aligned}
2^{3^{n+1}} 
&=
2^{3^n\cdot 3} = \left(2^{3^n} \right)^3\\
&\overset{P(n)}{=\!=\!=}
\Bigg[\ -1 + 3^{n+1}(1+3N(n))\ \Bigg]^3
\\
&=(-1)^3 
+3\cdot 3^{n+1}(1+3N(n))
\\
&\qquad\qquad\qquad
-\underbrace{3\cdot 3^{2(n+1)}}_{\equiv 0}(1+3N(n))^2
+ \underbrace{3^{3(n+1)}}_{\equiv 0}(1+3N(n))^3
\\
&\equiv 
-1 + 3^{n+2}(1+3N(n))\qquad\text{ modulo }3^{n+3}
\ .
\end{aligned}
$$
(We have used $n+3\le 1+2(n+1)\le 3(n+1)$, which is valid for $n\ge 1$.)
The claim is now inductively shown.
$\square$

Alternatively, note that $3^n$ is odd, use
$$
1+2^{3^n}=1+(\color{brown}3-1)^{\color{red}{3^n}}\\
=1+(-1) 
+ \binom{\color{red}{3^n}}1\cdot \color{brown}3^1 
- \binom{\color{red}{3^n}}2\cdot \color{brown}3^2 
+ \binom{\color{red}{3^n}}3\cdot \color{brown}3^3
-\dots
\pm\binom{\color{red}{3^n}}k\cdot \color{brown}3^k
\mp\dots
$$
and show (inductively) that the terms corresponding to $k\ge 2$ are each divisible by $3^{n+2}$.
