Given an odd $x$ there is an $m,n$ such that $2^n + 1 = 3^m x$? I'm curious about this question: Is it true that for any odd number $x\in 2\mathbb N + 1$ there exists numbers $m,n\in \mathbb N \cup \{0\}$ such that 
$$2^n+1 = 3^mx$$
Edit: I'm not trying to make this over-complicated and maybe there is somethng easier than what I am thinking. But, if Pillai's conjecture is true then the answer should be negative for most $x$.
 A: $7$ is one such number than cannot be expressed as $(2^n+1)/3^m$. For otherwise $3^m\cdot7=2^n+1$ for some $m,n$ positive integers. Thus $2^n+1=0 \mod 7$. But $2^n$ cannot have in $\mod 7$ values other than $2,4,1$ and so $2^n+1$ can have values only $2,3,5$ and none of these are $0\mod 7$.  
A: Here are some$\newcommand\leg[2]{\left(\frac{#1}{#2}\right)}\newcommand\ifs{\text{if }}\let\v\nu$ criteria. Either way, I don't expect there to be a simple characterisation of such $x$.


*

*If $m>0$, then for $2^n+1$ to be a multiple of $3$, $n$ should be odd. From $2^n+1\equiv0\pmod x$ we have $(2^{(n+1)/2})^2\equiv-2\pmod x$, so $-2$ is a quadratic residue modulo $x$. Because
$$\leg{-2}p=\leg{-1}p\leg2p=\begin{cases}-1&\ifs p\equiv5,7\pmod8\\1&\ifs p\equiv1,3\pmod8\end{cases}$$
we obtain that every prime divisor of $x$ is $\equiv1,3\pmod8$. (Because $3^2\equiv1\pmod8$ this implies in particular that $x\equiv1,3\pmod8$.)
If $m=0$ we simply have $x=2^n+1$ as a necessary and sufficient condition.

*By Zsigmondy's theorem, if $n$ has $\tau(n)$ positive divisors, then $2^n+1$ has at least $\tau(n)$ distinct prime divisors (unless $n=3$). This lets us conclude for example that $x$ cannot be a power of $3$ (though you might prove this using more elementary methods) unless for $x=1,3,9$. Also, if $x$ is given, Zsigmondy gives an upper bound on the number of prime divisors of $n$.

*If $m>0$ the Lifting The Exponent gives the exact exponent of $3$ in the factorisation of $n$:

Lemma. (Lifting The Exponent, special case)
  Let $p$ be an odd prime and $a,b\in\mathbb Z$ such that $p\mid a+b$ but $p\nmid a$. If $k$ is odd, then $\v_p(a^k+b^k)=\v_p(a+b)+\v_p(k)$, where $\v_p(k)$ denotes the exponent of $p$ in the prime factorisation of $k$.

So if $m>0$, then $n$ is odd hence $\v_p(n)=m-1$ by LTE. Combined with Zsigmondy this gives some more interesting partial results, such as: if $x=p^k$ is a prime power (with $p>3$ because $p=3$ is impossible, see above), then $n$ is prime. If $n>3$, then by LTE we have $m=1$, so $2^n+1=3p^k$.
A: For $x > 8$, $n \geq 3$, so $2^n + 1 = 1 \mod 8$.
$3^m = \begin{cases}3 \mod 8 & \text{ if } m \text{ odd} \\
1 \mod 8 & \text{ if } m \text { even}\end{cases}$
Looking at the equation $\mod 8$, we must therefore have
$1 = 3x \mod 8$ or $1 = x \mod 8$
So any solution must have $x = 1 \mod 8$ or $x = 3 \mod 8$, since the inverse of $3 \mod 8$ is $3$. So there are no solutions with $x > 8$ and $x = 5 \mod 8$ or $x = 7 \mod 8$ .
