Suppose that $m \geq 3$, $\mathcal{F}$ is the free group whose basis is $\{a,b\}$, and H is the normal subgroup of $\mathcal{F}$ generated by $\{bab^{-1}a, a^{2^{m-1}}, b^{-2}a^{2^{m-2}}\}$. Let $G_m = \mathcal{F}/H$. I'm interested in computing the order of $G_m$.
Given the above, I believe we have the presentation:
$G_m = \langle a,b \;|\; bab^{-1}a = a^{2^{m-1}} = b^{-2}a^{2^{m-2}} = e \rangle$.
Since $a^{-1} = a^{2^{m-1} - 1}$, $ba = a^{-1}b$, and $b^4 = e$ (because $a^{2^{m-1}} = b^{-2}a^{2^{m-2}} \implies b^{-2} = a^{2^{m-2}}$, and $b^{-2}a^{2^{m-2}} = e \implies b^{2} = a^{2^{m-2}}$), I think this means that every element in $G_m$ can be expressed in the form $a^ib^j$ or $b^ja^i$ with $i,j \geq 0$. Given that $a^{2^{m-1}} = e$ and $b^4 = e$, this implies that $|G_m| \leq 2(4 \cdot 2^{m-1})$. However, this is only an upper bound, and I'm looking to compute the exact order of $G_m$ for any choice of $m \geq 3$. I think that $G_3 = Q_8$, the quaternion group, so I'd conjecture that $|G_m| = 2^m$, but I'm not sure how to prove this.
Your group is a special case of the dicyclic group (replace $n$ by $2^{m-2}$): $$G=\langle a,b \mid a^{2n} = 1 , x^2 = a^n, xax^{-1}=a^{-1} \rangle$$ You already shown that $|G|\leq 4n$. It suffices to find a group $K$ with order $4n$ which satisfies the above relations: the subgroup of $\mathbb{H}^{\times}$ (non-zero quaternion under multiplication) generated by $a=e^{i\pi/n}, b=j$ does the job. It is not hard to verify $|K| = 4n$.