Here is a proof which relies on a straighforward generalization of the Ping Pong Lemma.
Claim 1 Let $a$ and $b$ be the transformations of the Riemann sphere $\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$ defined by $ a(z) = \frac{1}{z} + 2,\quad b(z) = z + 4. $ Let $d \in \mathbb{N}_{> 0} \cup \{\infty\}$. Then the subgroup of $\operatorname{PGL}(2, \mathbb{Z})$ generated by $X_d = \{b^{k}ab^{-k} \,\vert\, 0 \le k < d\}$ is free on $X_d$.
Claim 1 and its proof are a trivial adaptation of [1, Example II.B.26] which makes use of Schottky groups.
Proof of Claim 1. Let $\mathbb{D} = \{ z \in \mathbb{C} \, \vert \, \vert z \vert \le 1\}$, $\Omega_k = (\mathbb{D} + 4k) \cup (\mathbb{D} + 4k + 2)$ for $0 \le k < d$. Observe that $\gamma_k = b^ka b^{-k}$ maps
- the exterior of the disk $\mathbb{D} + 4k$ onto the interior of the disk $\mathbb{D} + 4k + 2$, and
- the interior of the disk $\mathbb{D} + 4k$ onto the exterior of the disk $\mathbb{D} + 4k + 2$.
Thus $\gamma_k^m (\Omega_l) \subseteq \Omega_k$ for every $k \neq l$ and every $m \in \mathbb{Z} \setminus \{0\}$. Clearly $\Omega_k \nsubseteq \Omega_l$ if $k \neq l$. A straightforward generalization of the Table-Tennis Lemma yields the result.
Here is another way to answer OP's question by means of a generalized Table-Tennis Lemma. The following is a geometric abstraction wrapping the solution that HJRW has outlined in comments.
Claim 2 [1, Solution of Exercise II.b.36.i]. Let $\gamma_1, \dots,\gamma_d$ hyperbolic automorphisms of a tree $T$ with pairwise disjoint axes.Then the subgroup of $\operatorname{Aut}(T)$ generated by $\gamma_1, \dots, \gamma_d$ is free on these elements.
An automorphism $\gamma$ of a tree $T$ is hyperbolic if it acts on $T$ without inversion, i.e, it doesn't swap the ends of an edge of $T$, and if $\tau(\gamma) = \min_{x \in V(T)} {\bf d}(x, \gamma(x)) > 0$ where $\bf{d}$ is the usual combinatorial distance on $T$ (i.e., the reduced edge-path length) and $V(T)$ denotes the set of vertices of $T$. If $\gamma$ is hyperbolic, then the vertices $x$ satisfying ${\bf d}(x, \gamma(x)) = \tau(\gamma)$ are the vertices of a subgraph of $T$ which is a geodesic line called the axis of $\gamma$. We denote this line by $\operatorname{axis}(\gamma)$.
Proof of Claim 2. Re-indexing the elements $\gamma_i$ if need be, we can find $x_i \in \operatorname{axis}(\gamma_i)$ for every $i \in \{1, \dots, d\}$ such that the geodesic segment joining $\gamma_1$ to $\gamma_d$ passes through $\gamma_2, \dots, \gamma_{d - 1}$, in this order. Let $\Omega_i$ be the set of vertices of $T$ such that geodesic segment joining $x$ to $x_i$ passes through $\gamma_i(x_i)$ or $\gamma_i^{-1}(x_i)$. It is not difficult to check that $\gamma_i(\Omega_j) \subseteq \Omega_i$ and $\Omega_i \nsubseteq \Omega$ if $i \neq j$. Thus the generalized Table-Tennis Lemma applies.
Let us observe that Claim 2 also answers OP's question. Indeed, the element $\gamma_i = b^i a b^{-i}$ is a hyperbolic automorphism of the Cayley graph of the free group with basis $\{a, b\}$ and we have $\operatorname{axis}(\gamma_i) = \{b^i a^k \, \vert \, k \in \mathbb{Z}\}$, so that Claim 2 applies.
[1] P. de la Harpe, "Topics in Geometric Group Theory", 2000.