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While writing this answer I got an idea to generalize the problem.

Question. Could anybody tell me such $(a,b,n,k) \in \mathbb{N}^4$ $4$-tuple, $a,b,n,k>1$, $a \neq b$, which satisfies the following equality:

$$r(a^n) = b^k,$$

where $r$ denotes the reverse of a number.

user153012
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    $r(12^2) = 21^2$ and $r(13^2) = 31^2$ – Michael Albanese Sep 14 '14 at 01:06
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    See also https://oeis.org/A035123 "Roots of 'non-palindromic squares remaining square when written backwards'." 12, 13, 21, 31, 33, 99, 102, 103, 112, 113, 122,.... and the less interesting list at https://oeis.org/A035125, Roots of 'non-palindromic cubes remaining cubic when written backwards', where all 24 entries listed have only zeros and ones for digits. – Gerry Myerson May 01 '16 at 06:41

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One interesting example is $r(2201^3)=2201^3$, see https://oeis.org/A002780.

Gerry Myerson
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