Can two different natural numbers raised to the same irrational power be integers?

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Is it possible to find an $\alpha\in\mathbb R$, $\alpha>0$ so that for all $n\in\mathbb N$, $n^\alpha\in\mathbb N$?

A stronger(restricted) problem: Can $2^\alpha$ and $3^\alpha$ be simultaneously integers?? Here $\alpha >0$.

For the second one, I can only reduce to $\cfrac{\log 3}{\log 2} = \cfrac{\log s}{\log q}$ for integer solution $(s,q)$, and then no idea..... For the first one, we have excluded all $\alpha\in\mathbb Q$(which is easy). I tried to use large enough $n^\alpha\in\mathbb N$ and to show $(n+1)^\alpha$ fails to be an integer but I failed.

I think there is more advanced number theoretic technique to be used.

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Let $$\alpha = log _3 7$$ Then

$$ 3^\alpha ,9^\alpha, 27^\alpha, 81^\alpha,.......$$

are all integers.