Question as in the title : does anyone know how to prove that $3^n$ does not divide $4^n+5$ for $n\geq 2$ or find a counterexample ?

My thoughts : (1) I have checked that this is true for $n\leq 1000$.

(2) I asked a similar [question](https://math.stackexchange.com/questions/3626676/3n-does-not-divide-8n1-for-n-geq-4) recently, and it was successfully solved with a method that uses a "lifting exponent lemma" which ultimately reduces to the identity $x^k-y^k=(x-y)(x^{k-1}+x^{k-2}y+\ldots+y^{k-1})$. Since $4^n+5$ cannot be so factored, this does not seem to apply here.

(3) For $r\geq 0$, denote by $q\_r$ the smallest positive integer such that $4^{q\_r}+5$ is divisible by $3^r$. It is easy to see that the order of $4$ modulo $3^r$ is exactly $3^{r-1}$, and hence $3^r$ divides $4^n+5$ iff $n\equiv q\_r \ \pmod{3^{r-1}}$. It follows that $q\_{r+1}\equiv q\_r \ \pmod{3^{r-1}}$ and so we have a decomposition in base three, $q\_r=\sum\_{j=0}^{r-1}\varepsilon\_j 3^j$ (where $\varepsilon\_0=q\_0$ and $\varepsilon\_k=\frac{q\_k-q\_{k-1}}{3^{k-1}}\in\lbrace 0,1,2\rbrace$ for $k\geq 1$). The first terms of the $\varepsilon$ sequence are

$$
\varepsilon\_0=1,\varepsilon\_1=2, 2, 1, 1, 0, 0, 2, 1, 0, 0, 0, 1, 1, 2, 0, 0, 1, 2
$$
No pattern seems to emerge at this point.