Solution:

Let $Y$ be the projection of $X$ onto $AB$. We prove that the circles circumscribed to triangles $AMC$ and $AXH$ both pass through $Y$. This is equivalent to proving that $BY \cdot BA = BM \cdot BC$, and this in turn is true since both products equal $BX \cdot BH$, since the quadrilaterals $AYXH$ and $HXMC$ are cyclic, both having by construction a pair of opposite right angles.

At this point the thesis has become that the points $Y, Q, C$ are collinear if and only if $BQ$ and $CQ$ are perpendicular.

Let us now denote by $\theta$ the measure of angle $BAM$. From the cyclicity of $AYMC$ we know that $\angle YCB = \theta$. Moreover $\angle QBC = 90^\circ - \theta$, since in the isosceles triangle $BQM$ the angle at the vertex $Q$ has measure $2\theta$ (here we are using that $Q$ is the circumcenter of $ABM$ and central angles are twice the inscribed angles). It follows that $Y, Q, C$ are collinear if and only if $\angle BCQ = \angle BCY = \theta$, that is, if and only if $\angle BCQ + \angle QBC = 90^\circ$, that is, if and only if $BQ$ and $CQ$ are perpendicular.

Remark A posteriori, in the configuration in which $BQ$ and $CQ$ are perpendicular, the triangle $BQM$ turns out to be equilateral. It follows that $\angle BQM = 60^\circ$, and hence $\theta = 30^\circ$.