It sure feels like something that should be related to Schanuel's conjecture. Note that this is equivalent to finding integers $m$ and $q$ such that $$x = \frac{\ln m + 2\pi i q}{\ln 3}$$ is not an integer but $2^x$ is. [$3^x$ simplifies to $m$ by construction].
Continuing this, let us actually compute $2^x$. If we split it into real and imaginary parts, a large messy expression ensues. But it naturally splits into 2 reasonable cases, depending on whether $m\gt 0$ or $m \lt 0$. Let's deal with the positive case first. We get $$2^x = m^{\log_3 2}\left( \cos(2q\log_3(2) \pi) + i \sin(2q\log_3(2) \pi)\right)$$ For that to be an integer, it has to at least be real, but unless $2\log_3(2)q$ is an integer, the $\sin$ term will not be $0$. For $q=0$, and this is $m^{\log_3 2}$. We can rewrite that as $2^{\log_3 m}$. But we assumed that $m$ was a power of 3, (say $3^k$) then $m^{\log_3 2}$ simplifies to 3$, so$2^y$. But then \log_3 m$ (and thus $x$ x$) is an integer. For the$m \lt 0$, we get the slightly more complicated $$2^x = (-m)^{\log_3 2}\left( \cos(2(2q+1)\log_3(2) \pi ) + i \sin(2(2q+1)\log_3(2) \pi )\right)$$ Since$m\lt 0$, the first term is real, so we need$2 (2q+1)\log_3(2)$to be an integer for the$\sin$term to disappear, which cannot happen. Can someone find a flaw in my reasoning? I somehow expect so, as this did not seem difficult, and I would expect it to be if it's an open problem! 2 expand answer into what seems like a solution Continuing this, let us actually compute$2^x$. If we split it into real and imaginary parts, a large messy expression ensues. But it naturally splits into 2 reasonable cases, depending on whether$m\gt 0$or$m \lt 0$. Let's deal with the positive case first. We get$$2^x = m^{\log_3 2}\left( \cos(2q\log_3(2) \pi) + i \sin(2q\log_3(2) \pi)\right)$$For that to be an integer, it has to at least be real, but unless$2\log_3(2)q$is an integer, the$\sin$term will not be$0$. For$q=0$, and$m$a power of 3, (say$3^k$) then$m^{\log_3 2}$simplifies to$2^y$. But then$x$is an integer. For the$m \lt 0$, we get the slightly more complicated $$2^x = (-m)^{\log_3 2}\left( \cos(2(2q+1)\log_3(2) \pi ) + i \sin(2(2q+1)\log_3(2) \pi )\right)$$Since$m\lt 0$, the first term is real, so we need$2 (2q+1)\log_3(2)$to be an integer for the$\sin\$ term to disappear, which cannot happen.