Answer: $\angle MXN = 2\angle ABM = 46^\circ$.

Let $\angle BAC = 2\alpha$. The triangles $ABM$ and $ACN$ are similar, therefore $\angle CNL = \angle BML = \alpha + 23^\circ$. Let $K$ be the midpoint of the arc $BC$ of the circumcircle of the triangle $ABC$. Then $K$ belongs to the line $AL$ and $\angle KBC = \alpha$. Both $X$ and $K$ belong to the perpendicular bisector of the segment $BC$, hence $\angle BXK = \frac{1}{2}\angle BXC = \angle BML$, so the quadrilateral $BMXK$ is inscribed. Then

$$\angle XMN = \angle XBK = \angle XBC + \angle KBC = (90^\circ - \angle BML) + \alpha = 90^\circ - (\angle BML - \alpha) = 67^\circ.$$

Analogously we have $\angle CXK = \frac{1}{2}\angle BXC = \angle CNL$, therefore the quadrilateral $CXNK$ is inscribed also and $\angle XNM = \angle XCK = 67^\circ$. Thus, the triangle $MXN$ is equilateral and

$$\angle MXN = 180^\circ - 2 \cdot 67^\circ = 46^\circ.$$