Solution:

Solution I. We first prove that $\angle A Z X = \angle Y Z A$. Since the triangles $\triangle B A X$, $\triangle C A Y$ are isosceles, and $X Y$ is the external bisector of $\angle B A C$, we see that
$$\angle B X A = \angle X A B = \angle C A Y = \angle A Y C.$$
Using these equalities, we find that
$$\angle X B A = 180^\circ - \angle B A X - \angle A X B = 180^\circ - \angle C Y A - \angle Y A C = \angle A C Y$$
Since $B, C$ are the centres of respectively $\Gamma_{1}, \Gamma_{2}$, this implies that
$$\angle A Z X = \frac{1}{2} \angle A B X = \frac{1}{2} \angle Y C A = \angle Y Z A.$$
Let $O$ be the circumcentre of $\triangle X Y Z$, next we will prove that $\angle B Z O = \angle A Z Y$. Consider the configuration where $\angle Z X A$ is sharp, then $\angle A B Z = 2 \angle A X Z = 2 \angle Y X Z = \angle Y O Z$. Since $\triangle B Z A$ and $\triangle O Z Y$ are isosceles, this implies $\angle B Z A = \angle O Z Y$. Subtracting $\angle O Z A$ (or adding, depending on the configuration) yields $\angle B Z O = \angle A Z Y$.

Together with the analogous result $\angle C Z O = \angle A Z X$, we conclude $\angle B Z O = \angle A Z Y = \angle X Z A = \angle O Z C$, so $O$ lies indeed on the internal bisector of $\angle B Z C$.

Solution II. Let $O$ be the circumcentre of $\triangle X Y Z$, then we see that $O X = O Z$. Since $B$ is the centre of $\Gamma_{1}$, we also see that $B X = B Z$, so $O B$ is the perpendicular bisector of $X Z$. Therefore $\angle B Z O = \angle B X O$, and analogously we find $\angle C Z O = \angle C Y O$. Note that $B, C$ lie on the same side of $X Y$; we will consider the configuration where $O$ is on the opposite side of $X Y$. Then $\angle B X O = \angle B X A + \angle A X O$. Now the isosceles triangles $\triangle B X A, \triangle C A Y, \triangle O X Y$, and $X Y$ being the external bisector of $\angle B A C$ give
$$\begin{aligned} & \angle B X A = \angle X A B = \angle C A Y = \angle A Y C, \\ & \angle A X O = \angle Y X O = \angle O Y X = \angle O Y A, \end{aligned}$$
so we find that $\angle B Z O = \angle B X O = \angle A Y C + \angle O Y A = \angle O Y C = \angle O Z C$. Hence $O$ lies on the internal bisector of $\angle B Z C$.