In this question $F_{n}$ is the free group with $n$ generators.
Is there a subgroup of $F_{2}$, isomorphic to $F_{4}$, which index is finite but not in the form of $3k$(not multiple of $3$)?
The motivation: using covering space in algebraic topology we can embed $F_{4}$ in $F_{2}$ with index $3$.(see page 58, Algebraic topology A. Hatcher). On the other hand there is a k index embeding of $F_{2}$ into $F_{2}$, for every $k\in \mathbb{N}$. The composition of this embeding is a $3k$ index embedding of $F_{4}$ in $F_{2}$.(So the question ask, "Is $3k$ the only possible index"?)
$\DeclareMathOperator{\rank}{rank}$ Using the formula $$[F:H]= \frac{\rank{H}-1}{\rank{F}-1}$$ where $F$ is a free group of rank finite and greater than 1, we can get that $[F_2:F_4]=(4-1)/1=3$, so the answer to your question is no and in fact $k$ must be $1$. You can find a proof of the formula in Combinatorial Group Theory, by Lyndon and Schupp, Prop3.9 Chapter 1. Also the formula can be found in Combinatorial Group Theory by Magnus,Karrass, and Solitar, Theorem 2.10. Also note the existence of such a subgroup can be guarrenteed by a surjection $f: F_2 \to C_3$, and note that $\ker f$ is such a group.