The quaternion-brace counting conjecture

For $m\in\mathbb{N}$, let $Q_{4m}$ be the generalized quaternion group, and call a brace a quaternion brace if its multiplicative group is isomorphic to a generalized quaternion group. Let $q(4m)$ be the number of isomorphism classes of quaternion braces of size $4m$. Quaternion-brace counting conjecture. For $m>2$,

$$q(4m)=\begin{cases} 2 & \text{if } m\text{ is odd},\\ 7 & \text{if } m\equiv0\bmod8,\\ 9 & \text{if } m\equiv4\bmod8,\\ 6 & \text{if } m\equiv2\bmod8\text{ or }m\equiv6\bmod8. \end{cases}$$

The authors checked the conjecture for all $m\leq512$. No proof or refutation beyond that finite verification is supplied, so the general statement remains open.