Let $2^n-13^m=k^3$ for some $m,n,k\in\mathbb{N}$. We have $2^n\in\{1,5,8,12\}\pmod{13}$, so $3\mid n$. Set $n=3t$ and factorize as
\[
\left(2^t-k\right)\left(2^{2t}+2^t\cdot k +k^2\right) = 13^m.
\]In particular, for some $\alpha,\beta\ge 0$ with $\alpha+\beta =m$, we have $2^t-k=13^\alpha$ and $2^{2t}+2^t\cdot k+k^2=13^{\beta}$. If $\alpha>0$ then $2^t\equiv k\pmod{13}$. So, $2^{2t}+2^t\cdot k +k^2\equiv 3\cdot 2^{2t}\not\equiv 0\pmod{13}$, a contradiction. Thus, $\alpha=0$ and $2^t=k+1$. Plugging this in, we obtain
\[
3\cdot 2^t\left(2^t-1\right) = 13^m-1.
\]Now, it is not hard to check that for $t=3$, $m=2$ is a solution, yielding $(m,n,k)=(2,9,7)$. Suppose $t\ge 4$. Using lifting the exponent (or direct factorization), we have $v_2(13^m-1) = v_2(m)+2=t$, so $v_2(m)=t-2$. Hence, $2^{t-2}\mid m$. Moreover, $2^{t-2}\ge t$ for $t\ge 4$, so
\[
2^{2t+2}>3\cdot 2^{2t} = 3\cdot 2^t -1 +13^m > 13^{t}>2^{3t} \Rightarrow t<2,
\]a contradiction.