$\DeclareMathOperator\rank{rank}$Let $F$ be a non-cyclic free group.

For which finitely generated subgroups $H< F$ such that $H$ is not of finite index in a free factor of $F$ does there exist a two-generated proper subgroup $K=\langle x, y\rangle< F$ such that $H\leq K$?

I'd be happy with a proof that it holds for all such finite subsets (unlikely?), and satisfied by a single counter-example. But a nice description of all these subgroups $H$ would be ideal.

The restriction on $H$ not being of finite index in $F$ is clearly necessary, as if $H$ has finite index in $F$ and is contained in a two-generated subgroup $K=\langle x, y\rangle$ of $F$, then we necessarily have $K=F$ (by for example the Schreier index formula). [For a similar reason, if $F$ contained a maximal subgroup $M$ which is finitely generated and not two generated then this would yield a counter-example. However, maximal subgroups of $F$ are not finitely generated.]

More generally, finite index subgroups of free factors have been removed from the question as these can also give counter-examples. For example, if $F=F'\ast F''$ with $F'$ of rank at least $4$ and $H$ has index $2$ in $F'$, then $H$ is not contained in any two-generated subgroup $K$ of $F$ (as otherwise $F=\langle H, a, F''\rangle=\langle K, a, F''\rangle$ where $F'=H\cup Ha$, and so $F$ could be generated by $2+1+\rank(F'')<\rank(F')+\rank(F'')=\rank(F)$ elements).