Let $W$ be the midpoint of the major arc $BAC$, $A' \in (ABC)$ be such that $AA' \parallel BC$, $T$ be the intersection of the circle with diameter $AI$ and $(ABC)$ and let $\omega$ touch $AC$, $AB$ at $E$, $F$. We claim the two circles touch at $T$.

Firstly, observe that $Z \in (AEF)$, so
$$\begin{aligned} \angle ATZ &= \angle AIZ = 90^\circ - \left(\frac{\alpha}{2} + \gamma\right) \\ &= \frac{\beta - \gamma}{2} = \beta - \left(90^\circ - \frac{\alpha}{2}\right) \\ &= \angle ABC - \angle WBC = \angle ABW = \angle ATW, \end{aligned}$$
hence $T$, $Z$, $W$ are collinear. Let $TW \cap BC = P$. By spiral similarity and angle bisector theorem we have $\frac{PB}{PC} = \frac{TB}{TC} = \frac{BF}{CE}$, so by converse of Menelaus theorem for $\triangle ABC$ we obtain that $EF$, $TZ$, $BC$ are concurrent at $P$. Thus, by power of point at $P$, we obtain that $PX \cdot PY = PE \cdot PF = PZ \cdot PT$, so $XYZT$ is cyclic.

Finally, observe that the center of $(ZXY)$ lies on $IZ$, so $(ZXY)$ touches $AA'$ at $Z$. Hence, by shooting lemma, since $W$ is the midpoint of the minor arc $AA'$, we obtain that $(ZXY)$ touches $(ABC)$ at $T$ and we are done. $\square$