The number $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$.
Let $$A_n = 2^{3^n}+1.$$ The claim holds for $n=0$, and since: $$ A_{n+1} = (A_n-1)^3+1 = A_n^3 -3A_n(A_n-1) $$ we have that $\nu_3(A_{n+1}) = 1+\nu_3(A_n)$, since $9\mid A_n^3$ and $3\mid 3A_n$ but $3\nmid (A_n-1)$.