Mariano's answer is quite right, but relies on the 'accident' that all subgroups of index two are normal. If you wanted to find all the subgroups of index three, say, you would need to try another approach. Here's one idea.
The free group of rank two, $F$, is the fundamental group of the figure-eight graph $\Gamma$, which has one vertex and two edges. Fix the vertex as the base point. The subgroups of index $k$ correspond, via covering-space theory, to connected covering spaces $\widehat{\Gamma}\to\Gamma$ of degree $k$, with a choice of base vertex $\hat{v}\in\widehat{\Gamma}$. (If, on the other hand, you only wanted to count conjugacy classes of subgroups, you could forget about $\hat{v}$.) Because the covering map $p:\widehat{\Gamma}\to\Gamma$ has degree $k$, the graph $\widehat{\Gamma}$ has exactly $k$ vertices.
Let's decorate the graph $\Gamma$ so that we can see $F=\langle a,b\rangle$ more clearly. Label each edge with the corresponding generator $a$ or $b$. Furthermore, orient each edge to indicate in which direction the generator goes around it.
You can use the covering map $p$ to pull the labels and orientations back from $\Gamma$ to $\widehat{\Gamma}$. That is, if $\hat{e}$ is an edge of $\widehat{\Gamma}$ and $p(\hat{e})=e$ is labelled $a$, then you should label $\hat{e}$ with $a$ also, and orient $\hat{e}$ so that $p$ sends the orientation on $\hat{e}$ to the orientation on $e$.
With a little thought, it's not too hard to see that this decoration on $\widehat{\Gamma}$---the labels and orientation---are enough to reconstruct the map $p$: they tell you where to send each edge, and there's only one choice of where to send each vertex. The statement that $p$ is a covering map translates into a nice condition on the decoration of $\widehat{\Gamma}$: for each vertex $\hat{u}$, you see exactly one edge with label going into $\hat{u}$, and exactly one edge with each label going out of $\hat{u}$. I'll call this condition 'the covering condition'.
This discussion turns counting the subgroups of index $k$ into the combinatorial problem of counting the connected, decorated, based graphs $\widehat{\Gamma}$ with $k$ vertices that satisfy the covering condition. With a little thought, one can write down a formula for this. Marshall Hall Jr did just this (though I don't think he thought in terms of the covering spaces of graphs) and came up with the following formula. Let $N(k,r)$ be the number of subgroups of index $k$ in the free group of rank $r$. Then
$N(k,r)=k(k!)^{r-1}-\sum_{i=1}^{k-1}((k-i)!)^{r-1}N(i,r)$.
For some related things, I wrote a blog post about proving theorems about free groups by thinking about covering spaces of graphs here.
Alternatively, if you don't like covering spaces, an equivalent point of view is to count transitive actions by permutation groups with $r$ generators on based sets of size $k$. This turns out to be the same combinatorial problem.
If $F_3$ were a quotient of $F_2$, then $\mathbb{Z}^3$ would be, but $\mathbb{Z}^3$ cannot be generated by fewer than $3$ elements. To me it seems easier to see directly that $\mathbb{Z}^3$ needs at least $3$ generators than the corresponding statement for $F_3$, perhaps because it's easy to visualize.
The rank of a group is the smallest cardinality of a generating set. Here's a list of some facts about ranks of groups (including that the rank of $F_3$ is $3$) on Wikipedia.
Best Answer
I would do this computation by thinking about covering spaces of graphs and applying Stallings's folding algorithm.
We realise $F_2$ as the fundamental group of a graph $X$ with one vertex $v$ and two edges---this is sometimes called the rose with two petals. We orient the edges, and label one by $a$ and the other by $b$. This fixes an identification $\langle a,b\rangle\equiv\pi_1(X,v)$.
Your subgroup $H$ can be thought of similarly. It is the fundamental group of a graph $Y$, which we construct as follows:
The orientations and labels define a natural map $Y\to X$, and the image of $Y$ is your subgroup $H$.
(For more on this sort of construction see, for instance, this blog post.)
Stallings' folding algorithm is a way of turning this map into an immersion---that is, a local embedding. The algorithm is easy:
At the end of this procedure, we have a new oriented, labelled graph $Y'$, and the map $Y'\to X$ is an immersion. There are essentially two possibilites:
Every vertex of $Y'$ has valence four. If so, then $Y'\to X$ is a covering map and $H$ is a finite-index subgroup of $F_2$. The index of $H$ is equal to the degree of the covering map, which is equal to the number of vertices of $Y'$.
Some vertex of $Y'$ has valence less than four. If so, then $H$ is of infinite index. (To see this, note that you can complete it to an infinite-sheeted covering space.)
In your case, you can quickly see that you need to perform exactly one fold to turn $Y$ into $Y'$. If $k=1$ then $Y'$ is isomorphic to $X$ and your subgroup $H$ is equal to $F_2$. Otherwise, $Y'$ has a vertex of valence two and $H$ is of infinite index.