If 13 does not divide m, then prove that $m^4+8$ is not a cube of an integer

Question

My question is how can we prove that $m^4 + 8$ not a cube of an integer if $m$ can not be divided by 13.

What I have done so far:

By Fermat’s Little Theorem: \begin{align} m^{p-1} &\equiv 1 \pmod p \\ m^{13-1} &\equiv 1 \pmod{13} \\ m^{12} &\equiv 1 \pmod{13} \\ (m^4)^3 &\equiv 1 \pmod{13}. \end{align}

I am not sure how to use this information to prove the original question though.