Let $F$ be free group on $\{x,y\}$. Consider the element $x^2$. Does there exists $z\in F$ such that $\{z,x^2\}$ generate whole $F$?
I thought in following direction. $x^2$ is not in commutator subgroup. If there is such a $z$ then $\{\overline{x}^2,\overline{z}\}$ will generate abelian group $F/[F,F]=\langle \overline{x}\rangle \oplus \langle \overline{y}\rangle$, but this is not possible, since from set $\{\overline{x}^2,\overline{z}\}$, we obtain only even powers of $\overline{x}$ in $F/[F,F]$. Is this correct?
More generally, given any $a=W(x,y)\in F$, what condition on $a$ guarantees that $\{a\}$ can be extended to a basis of $F$?
The Klein four-group $V=\{1,a,b,c\}$ is generated by $\{a,b\}.$ Define a surjective homomorphism $h:F\to V$ with $h(x)=a$ and $h(y)=b.$ If $F$ were generated by $\{x^2,z\},$ then $V$ would be generated by $\{h(x^2),h(z)\}=\{1,h(z)\},$ so $V$ would be cyclic, which is absurd.