What are the possible prime factors of $3^n+2$ , where $n$ is a positive integer? 
What are the possible prime factors of $3^n+2$, where $n$ is a positive integer ?

It is clear that a prime $p$ for which neither $-2$ nor $-6$ is a quadratic residue modulo $p$, cannot be a prime factor of $3^n+2$, so the primes congruent to $13$ or $23$ modulo $24$ can be ruled out. But is there a simple necessary and sufficient condition whether $p$ can be a prime factor of $3^n+2$ ?
 A: Just an observation not a complete answer.
For a prime $p>3$, applying Fermat's little theorem
$$2^p \equiv 2 \pmod{p}$$
and if there is a minimal $n_0 \in \mathbb{N}$ such that ($p$ - odd!):
$$3^{n_0} \equiv -2 \pmod{p} \Rightarrow 3^{n_0\cdot p} \equiv (-2)^p \equiv -2 \pmod{p}$$
Also $n_0<p-1$, otherwise $n_0=(p-1)\cdot q + r, 0\leq r < p-1$ and
$$3^{n_0}=3^{(p-1)\cdot q + r}=3^{(p-1)\cdot q}\cdot 3^{r}\equiv 3^r \equiv -2 \pmod{p}$$
since $3^{p-1}\equiv 1 \pmod{p}$. So, we can look for such mininal $n_0$ within $\{0,1,2,..,p-1\}$ range.
For example for


*

*$p=5$, $n_0=1$ because $3 \equiv -2 \pmod{5} \Rightarrow 3^{5^k} \equiv -2 \pmod{5}$.

*$p=7$, $n_0=5$ because $3^5 \equiv -2 \pmod{7} \Rightarrow 3^{5\cdot7^k} \equiv -2 \pmod{7}$.

*$p=11$, $n_0=2$ because $3^2 \equiv -2 \pmod{11} \Rightarrow 3^{2\cdot11^k} \equiv -2 \pmod{11}$.

*$p=13$ there is no such $n_0$

A: COMMENT.-What I give here IS AN ANSWER. However I post it as a comment  because maybe it is not as readers expect.
It is clear that $p\ne 2,3$ Let $p|(3^n+2)$ so one has in the prime field $\mathbb F_p$ the equality
$$3^n+2=0\iff3^n+3=1\iff3(3^{n-1}+1)=3^{p-1}\iff \color{red}{3^{n-1}+1=3^{p-2}}$$
In particular, $3$ must be the inverse of $3^{n-1}+1$ in the field $\mathbb F_p$
$$\text{ EXAMPLES}$$
$$►n=6\Rightarrow3^6+2=17\cdot43\Rightarrow3^{5}+1=244
=\begin{cases}6=3^{15}\text{ in }\mathbb F_{17}\\29=3^{41}\text{ in }\mathbb F_{43}\end{cases}$$
$$►n=12\Rightarrow3^{12}+2=11\cdot48313\Rightarrow3^{11}+1=177148
=\begin{cases}4=3^{9}\text{ in }\mathbb F_{11}\\32209=3^{48311}\text{ in }\mathbb F_{48313}\end{cases}$$
NOTE.-I have put $32209=3^{48311}\text{ in }\mathbb F_{48313}$ at the end only calculating $177148$ modulo the prime $48313$ getting $32209$ and giving by true (it should be true!) the equality. 
