Interesting. It turns out that $g$ must lie in the commutator subgroup $[F_2,F_2]$. The reason is that if it doesn't, then the abelianization of $g$ is nontrivial (the commutator subgroup is the kernel of the abelianization homomorphism). This can we used to show that the conjugacy class of $g$ is $\{g\}$. But this is absurd, because $F_2$ has the ICC property (infinite conjugacy class property).