First of all, I have read this similar question and am satisfied that the answers there prove the result I am interested in. That being said, I'm more interested in this particular approach than in the actual result, so I think it's reasonable to post this as a new question, even if the approach covered there can be considered a "better" answer to the underlying question.
Fix the notation $$F_k := \textrm{ the free group on } k \textrm{ generators}.$$ Let $n,m \in \mathbb{Z}^+$ and furthermore let $n > 1$. I claim that there exists a subgroup $G$ of $F_n$ with $$G \cong F_m.$$ To "prove" this statement (actually, I believe the argument is only a heuristic):
First, the statement is trivial if $m < n$, so assume $m > n$. For the cases $n = 2, 3$, to demonstrate that $F_n$ contains an isomorphic copy of a "larger" free group, write $$F_2 = \langle a,b \rangle; \ \ \ G = \langle aa,bb,ab \rangle \subset F_2; \ \ \ G \cong F_3 $$ and $$F_3 = \langle a,b,c \rangle; \ \ \ G = \langle ab,ac,bc,aa \rangle \subset F_3; \ \ \ G \cong F_4. $$
Finally, to demonstrate that when $n \geq 4$, $F_n$ contains an isomorphic copy of a "larger" free group, write $$F_n = \langle g_1, \dots, g_n \rangle.$$
We use the observation that there are no relations between the elements of the set $$B := \{g_i g_j : 1 \leq i < j \leq n \}.$$ This set has cardinality ${n \choose 2} > n$, so the group $$G = \langle B \rangle \subset F_n$$ is isomorphic to $F_{n\choose 2}$.
Now, I'm not sure if the observation in the last few lines is actually true, and I would really love a counterexample if one exists or a hint of how I might begin a proof of it. But, supposing it holds true, then this line of reasoning apparently tells us how to write a proof of the original statement for any particular $n >1$, $m >n$. What I'm not sure of is how to write a rigorous argument that encapsulates every case without making use of the statement "and now if we wanted to we could..." or some such.
