Claim: If $F(X_1) \cong F(X_2)$, then $|X_1| = |X_2|$.
Proof: Look at the sets $\text{Hom}(F(X_1), F_2)$ and $\text{Hom}(F(X_2), F_2)$ of group homomorphisms to the field $F_2$ with $2$ elements. These sets are vector spaces over $F_2$ with bases $X_1$ and $X_2$ respectively. Fixing an isomorphism $θ : F(X_1) → F(X_2)$, we have an isomorphism of $F_2$-vector spaces from $\text{Hom}(F(X_2), F_2)$ to $\text{Hom}(F(X_1), F_2)$ given by $φ \to φ ◦ θ$. Thus, their bases must have the same cardinality, which proves that $|X_1| = |X_2|$.
In the above proof, I don't understand why $X_1$ and $X_2$ are bases for the vector space. Isn't the vector space supposed to be the set of functions between the groups?
I recently encountered the same defective proof in Robinson's book (Exercise 2.1.7). As a matter of fact, the claim (for infinite sets $X_i$) cannot be proven using $\mathrm{Hom}(X_i,\mathbb{F}_2)$. To see this, note that $|\mathrm{Hom}(X_i,\mathbb{F}_2)|=|2^{X_i}|$, since by the universal property every map $X_i\to\mathbb{F}_2$ extends uniquely to a homomorphism $F(X_i)\to\mathbb{F}_2$. However, the statement $|2^{X_1}|=|2^{X_2}|\Longrightarrow |X_1|=|X_2|$ cannot be proven in ZFC set theory! (It is a version of the generalized continuum hypothesis). Nevertheless, Robinson gave a correct proof in 2.3.9.