Suppose I have a finitely presented group $G$ with two generators $x$ and $y$, with one relator $r$. And suppose the relator is a cyclically reduced word (in $x$ and $y$), i.e., every cyclic permutation of the word is reduced.
If $G$ turns out to be a free group, then must the relator have length at most two, in the sense that $r=x^a y^b$ (or $y^b x^a$) with $a,b \in \mathbb{Z}$?
No.
A primitive word in a free group $F$ is a word which forms part of a free basis for $F$. Any primitive word $r$ as your relator works (which can be seen via Tietze transformations), and these have arbitrary length. (This is in fact an "if and only if" condition, but I cannot think of an easy proof for the opposite direction.)
For example, $(ab)^nb$ is primitive in $F(a,b)$ for any $n$ , as $\{ab, (ab)^nb\}$ is a free basis, and so $\langle a,b\mid (ab)^nb\rangle$ is free for any $n$.
(Note that if a two-generator one-relator group $G$ is free then it is free of rank one, i.e. infinite cyclic, as $G$ necessarily surjects onto the infinite cyclic group, so is not free of rank zero, while the free group of rank two is Hopfian so $G$ cannot be free of rank $\geq2$.)