A result of Schützenberger on commutators and powers in free groups It is an old result of Schützenberger that in a free group, a basic commutator cannot be a proper power. A look at the original reference
M.-P. Schützenberger, Sur l'équation $a^{2+n} = b^{2+m}c^{2+p}$ dans un groupe libre, C. R. Acad. Sci. Paris 248 (1959), 2435–2436 (French).
quickly reveals that a lot of details are missing and some claims appear to be wrong.

Question: Let $F$ be a free group and $a,b,c \in F$ with $c \neq 1$ and $n \geq 2$. Why is $[a,b] \neq c^n$?

In particular, it would be nice to have a somewhat geometric proof of this apparently fundamental fact. There is an algebraic proof in 
G. Baumslag, Some aspects of groups with unique roots, Acta Math 104(3) (1960), 217–303.
as Lemma 36.4 but it is relies on various technical computations and is hard to grasp.
 A: There is a visualisation of Schützenberger's
observation based on a variant of the
car-crash lemma that says that

for any multiple motion
on a map on a closed oriented
surface $S$, the
number of points of complete collision is at least
$\chi(S)+\sum\limits_D(d_D-1)$,
where the summation runs over all
faces $D$ of the map
and $d_D$ is the number of cars moving around a face $D$.

See
[this self-advertisement]
for exact definitions.
So, if a non-identity commutator in a free group $F(x,y,\dots)$ is, e.g.,
a cube of a word $c$, we obtain a one-face map on a torus, where the
label of the face is $c^3$. Let three cars move around this face
counter-clockwise
with a
constant speed one edge per a minute; at a moment
$t$ each car moves along an edge labelled by the $[t]$th letter
of the word $c$ (where the integer part $[t]$
of $t\in\mathbb R$ is counted modulo the length of $c$).

The car-crash lemma asserts that
$$
\small
\pmatrix{
\hbox{the number of points}
\\\\ 
\hbox{of complete collision}
}
\geqslant
\begin{pmatrix}
\hbox{the Euler characteristic}
\\\\ 
\hbox{of the torus,}
\\\\
\hbox{i.e. 0}
\end{pmatrix}
+
\pmatrix{
\hbox{$d_D$, i.e. the number of cars}
\\\\
\hbox{moving around the unique face $D$,}
\\\\
\hbox{i.e. 3}
}
-1=2.
%}
$$
On the other hand, a collision inside an edge is impossible:

such a collision would imply that some letter of the word
$c$ is $x$ and $x^{-1}$ simultaneously. A similar argument shows
that a collision at a vertex cannot occur too. This contradiction
proves that a commutator cannot be a cube.
This approach was used in
[this self-advertisement],
[this self-advertisement],
and
[this self-advertisement]
to obtain generalisations of
Schützenberger's observation
and
Duncan--Howie's
theorem (mentioned by Henry Wilton).
A: The case $n=2$ (originally due to Lyndon) admits a very nice geometric argument: one notes that elements $a,b,c$ with $[a,b]=c^2$ lead to a map from the surface $\Sigma_{-1}$ of Euler characteristic -1 to a graph.  Pulling back midpoints of edges, one obtains essential, two-sided, simple closed curves on $\Sigma_{-1}$.  Pinching these, Euler characteristic shows that one of the resulting components must be a projective plane, and the result follows.
For $n> 2$, Duncan--Howie proved the stronger theorem that, for any non-trivial element $c$ of $F_2$, the stable commutator length of $c$ is at least $1/2$.  (In fact, by passing to covers, one can also deduce the $n=2$ case from the Duncan--Howie theorem.)  When you decode it, their proof is quite geometric, relying on defining a sort of combinatorial vector field on surfaces.
Louder and I gave a sort of generalization of Duncan--Howie's proof in our work on one-relator groups. Again, the idea is to count Euler characteristic using some sort of geometric data (we call it a stacking).    We don't spell it out explicitly, but you can use our ideas to give a geometric proof of the Duncan--Howie theorem, and in particular of the result that you want.
A: 
(a) On the free 2-step nilpotent group $L$ on $(u,v)$, the commutator $z=[u,v]$ is not a proper power.

Indeed, since $L/[L,L]$ is torsion-free, any root of $z$ should lie in $[L,L]$ and the latter is readily seen to be the cyclic group generated by $z$. (Remark: $L$ is isomorphic to the integral 3-dimensional Heisenberg group.)$\qquad\Box$
I'll use three other very classical facts, without proof: the hardest (b) has both geometric and combinatorial proofs (the simplest using that $\pi_1$ of graphs are free), while (c),(d) are elementary exercises ((c) follows from Malcev's more general results but is very easy by hand here).

(b) Subgroups of free groups are free.
(c) $L$ is Hopfian (all its surjective group endomorphisms are automorphisms).
(d) If $N$ is a nilpotent group and $M$ a subgroup generating $N$ modulo $[N,N]$ (that is, $M[N,N]=N$), then $M=N$.

This allows to prove the result.

In a free group $F$, if $a,b,c$ are two elements and $[a,b]\neq 1$ then $[a,b]$ is not a proper power $c^n$ of $c$.

By (a), we can suppose that $F$ is generated by $a,b,c$. If $d$ is the rank of $F$, it is also the rank of its abelianization, and the relation $[a,b]=c^n$ then forces $d\le 2$. Also $d\ge 2$ since $a,b$ don't commute. So $F$ is free on 2 generators (up to now this is exactly Baumslag's argument).
Hence $L=F/[F,[F,F]]$ is free 2-step nilpotent on 2 generators. Denote by $A,B,C$ the images of $a,b,c$ in $L$. Since $L/[L,L]$ is torsion-free and $C^n\in [L,L]$, we have $C\in [L,L]$. By (d), we deduce that $L$ is generated by $A,B$. So the endomorphism of $L$ mapping the two free generators to $A$ and $B$ is surjective. By (c) deduce that $L$ is free 2-step nilpotent on $(A,B)$. Since $[A,B]=C^n$, we can use (a) to get a contradiction. (Baumslag's conclusion of the argument also uses a 2-step nilpotent group, but I haven't followed as it's more technical)
