Are there positive integers $n$ and $m$ for which reversing the base-10 digits of $2^n$ yields $7^m$?
I've answered this question for powers of 2 and powers of 3 in the negative. Permutations of the digits of a number divisible by 3 yields another number divisible by 3 in base-10. This arises from the fact that the sum of base-10 digits of a number divisible by 3 is itself divisible by 3.
Thus since $2^n$ is not divisible by 3, reversing the digits can't yield another number divisible by 3, and hence no natural number power of 2 when reversed will be a natural power of 3.
I'm currently trying to put limitations on n and m by considering modular arithmetic. Suggestions for techniques in the comments would also be appreciated.