For instance see [1,2], where the XTR is explained. Since they reduce $GF(p^6)$ to $GF(p^2)$ by considering their traces as representations of the equivalence class $Tr_{\mathbb F_{^{6/3}}}(a^{p^{m\cdot 6/3}})$ it is considerable to generalize this to consider for any $k$ with $3\mid k$ the trace $Tr_{\mathbb F_{a^{p^{k/3}}}}(a^{p^{m\cdot k/3}})$.
So my question occures: Is it possible to extend this method to any power of $p$, even if $3\not\mid k$? How about the field, where powers of $2$ devides $k$, that means $\exists m\in\mathbb N$ maximal, such that $2^m \mid k$. Let $n=2^m$, can't I reduce $\mathbb F_{p^k}$ to $\mathbb F_{p^{k/n}}=\mathbb F_{p^2}$?
[1] https://en.wikipedia.org/wiki/XTR
[2] https://infoscience.epfl.ch/record/164527/files/NPDF-23.pdf