I'm trying to generalize the question Is there such an $x$ that both $2^{\frac{x}{3}}$ and $3^{\frac{x}{2}}$ are simultaneously rational?.
Do there exist $a,b \in \mathbb{Q}^+ \setminus \{1\}$ and $x \in \mathbb{R} \setminus \mathbb{Q}$ such that:
The original question seems very hard. This question could be just as hard, but maybe I missed something.
Negative answer follows from the four exponentials conjecture. Indeed, assuming $x$ and $\frac{\log a}{\log b}$ are irrational, pairs $1,x$ and $\log a,\log b$ are linearly independent. The conjecture then implies that one of $$\begin{align*} e^{1\log a} &=a\\ e^{1\log b} &=b\\ e^{x\log a} &=a^x\\ e^{x\log b} &=b^x\end{align*}$$ is transcendental. Hence, for $a,b$ rational, $a^x,b^x$ can't be both rational.
It should be noted, however, that the four exponentials conjecture is wide open, even though we have proven some very similar results. Indeed, even the question of whether integrality of $2^x,3^x$ implies rationality of $x$ (which is a special case of your question) is well-known and open.
