1. Identify the External Similitude Center:
Let $H$ be the external similitude center of circles $\omega_1$ and $\omega_2$. By Monge's Theorem, the external similitude center of two circles and the internal similitude center of the same two circles lie on the line joining their centers. Since $\omega_1$ and $\omega_2$ are internally tangent to $\Omega$ at points $A$ and $B$ respectively, $H$ lies on the line $AB$.

2. **Inversion with Pole $H$:**
Consider an inversion with pole $H$. This inversion will map the circles $\omega_1$ and $\omega_2$ to themselves because $H$ is their external similitude center. The inversion will also map the points $A$ and $B$ to themselves because they lie on the line $AB$.

3. Fixing the Small Circle:
The small circle that is internally tangent to $\omega_1$, $\omega_2$, and $A_1B_1$ at $Z$ is fixed by this inversion. This is because the inversion with pole $H$ interchanges $\omega_1$ and $\omega_2$ and fixes the small circle that is tangent to both.

4. Equality of Power of Point:
Since the inversion fixes the small circle, we have:
$$HX^2 = HZ^2 = HA \cdot HB$$
This implies that the circle with center $H$ and radius $HY$ is an Apollonian circle of $XAB$.

5. Apollonian Circle Property:
By the property of the Apollonian circle, the angles subtended by the points $A$ and $B$ at any point on the circle are equal. Therefore, we have:
$$\angle AXZ = \angle BXZ$$

Thus, we have proven that $\angle AXZ = \angle BXZ$.

$\blacksquare$