Suppose we are given $a, b \in F_2$, that happen to generate it. Then must they be free generators? That is, there is no non-trivial reduced word on $a^{\pm 1}, b^{\pm 1}$ defining the identity. Equivalently, $a, b$ satisfy the universal property defining $F_2$.
I tried attacking this the first way: suppose $u$ and $v$ are free generators, then if I take a word on $a$ and $b$ I can rewrite it in terms of $u$ and $v$. But how do I know that if I took a non-trivial reduced word on $a, b$ defining the identity, it does not become trivial after reduction with $u$ and $v$?
I also tried to use the universal property. Then we get that if $a$ and $b$ are not free, $F_2$ is isomorphic to one of its propert quotients. Maybe this is not the case for $F_2$ specifically, but it may be for other groups, so no contradiction in sight once again.
I know this is true because it is used in a class I am taking. To prove that $SL_2(\mathbb{Z})$ is not free, the professor just exhibited two generators and a non-trivial relation.
Your question is equivalent to the following:
Groups with this property are called Hopfian groups: see the Wikipedia article .
In particular, it is stated that $F_n$ is Hopfian for all $n$: proof.
Thus, the answer to your question is yes: every set of 2 generators of $F_2$ is free.