In Johnson's 'Topics in the Theory of Group Presentations', one can find this theorem after the definition of free groups using the universal property.
Theorem. Free groups of different ranks are not isomorphic.
Proof. Let $F$ be free on a subset $X$ with $|X|= \omega$, and let $G$ be any group. Then it is the burden of definition of free group by the universal property that the mappings $X \to G$ are in one-to-one correspondence with the homomorphisms $F \to G$. Thus, there are exactly $2^{\omega}$ homomorphisms from $F$ to $\mathbb{Z}_2$. Since this number is invariant under isomorphism, we see that $2^{\omega}$, and hence the rank $\omega$, is determined by the isomorphism class of $F$.
Isn't the author assuming that $2^{|X|}=2^{|Y|} \Rightarrow |X| = |Y|$?
I have seen this post over here, https://mathoverflow.net/questions/67473/equality-of-cardinality-of-power-set, which says that this statement is independent from ZFC. So is the proof wrong?
Let me prove this in a different way. Observe that a free group $F(X)$ on a set $X$ is a free product (coproduct in the category of groups) of copies of $\mathbb{Z}$ indexed over $X$. Now, there is an adjunction $(-)^{ab} \dashv i$ where $(-)^{ab}$ is the abelianization functor from the category Grp of groups into the subcategory Ab of abelian groups and $i : \textbf{Ab} \hookrightarrow \textbf{Grp}$ is the inclusion functor. Since $(-)^{ab}$ is a left adjoint, hence it preserves colimits and in particular, coproducts. Thus, $F(X)^{ab}$ is a coproduct indexed over $X$ in the category of abelian groups, that is, a direct sum of copies of $\mathbb{Z}^{ab} = \mathbb{Z}$ indexed over $X$ (in other words, a free abelian group/free $\mathbb{Z}$-module with basis $X$).
Now, assume that there is an isomorphism $F(X) \xrightarrow{\cong} F(Y)$ for sets $X$ and $Y$. Since a functor preserves isomorphisms, hence applying the abelianization functor $(-)^{ab}$ to the isomorphism and using the above argument shows that
$\oplus_X \mathbb{Z} \cong \oplus_Y \mathbb{Z}$
Now, $\mathbb{Z}$, being a commutative ring, implies that it has the invariant basis number property. In other words, any two bases of a free $\mathbb{Z}$-module have the same cardinality. Thus, $|X| = |Y|$.