Consider a subset $\mathcal{J}^*$ of $\left\{ {1,2,...,m} \right\}$ such that each element in $\mathcal{J}^*$ divides $m$ and $k'$ divides $k$ for all $k,k' \in \mathcal{J}^*$, $k' \le k$.
Let us denote by $\mathcal{S}_k$ ($k=1,2,...,m$) a subgroup of cardinality $2^k$ of the additive group of $\rm GF$($q$). We choose $\mathcal{S}_k$ as the unique subfield of cardinality $2^k$ (which is in particular a subgroup) if $k \in \mathcal{J}^*$. For the remaining values of $k$, we choose subgroups.
As an example, if GF($64$) then $\mathcal{S}_1$, $\mathcal{S}_3$ or $\mathcal{S}_1$ and $\mathcal{S}_2$ are valid choices for subfields.
Is there a choice of subgroups among $\mathcal{S}_k$ (i.e., apart from the choice of subfields indexed in $\mathcal{J}^*$) such that $\mathcal{S}_1 \subset \mathcal{S}_2 \subset .... \subset \mathcal{S}_m = {\rm GF}(q)$?
To summarize the comments, the answer is yes, there is such choice of subgroups. The idea is to consider the subfields as subspaces and then to extend the bases (as pointed out by J. Lahtonen).