Let $XM$ intersect $\omega$ again at $A_1$, then by Reim $AA_1//FM//BC$, so $X$ lies on the symmedian. Let $H$ be the orthocenter. It is well-known that $X'$ is the $A-$humpty point, which lies on $(AH)$. Also let $AA'$ intersect the circumcircle at $D$, then invert wrt $A'$ gives $-1=(A,X;B,C)=(D,S;C,B)$, so $S$ is the second intersection of $(AH)$ and $(ABC)$. Let the antipode of $A$ wrt $(ABC)$ be $P$, then $AX=AZ$ gives $\measuredangle XAP=\measuredangle PAZ$, therefore $\measuredangle HSX'=\measuredangle HAX'=\measuredangle DAV=\measuredangle XAP=\measuredangle PAZ=\measuredangle PSZ=\measuredangle HSZ$, so $S,X',Z$ are collinear.

Wonderful configuration We structure the proof in some Claims. Claim 1: $AX$ is the $A-$ symmedian. Proof: Let $EM$ intersect $(ABC)$ again at $K$. Note that $\angle FXE=90^\circ=\angle KXE,$ and so points $K,F,X$ are collinear. Therefore, by the Ratio Lemma, $\dfrac{BX}{XC}=\dfrac{BF}{FC} \cdot \dfrac{\sin \angle FXC}{\sin \angle FXB}=\dfrac{BF}{FC}=\dfrac{AB}{AC},$ and so quadrilateral $ABXC$ is harmonic, implying the desired result $\blacksquare$ Claim 2: $X'$ is the reflection of $X$ in $BC$. Proof: Note that $\angle BCX=\angle BAX=\angle CAV=\angle CXV,$ and so $BC \parallel XV,$ hence $X'$ is the reflection of $X$ in $BC$ $\blacksquare$ Claim 3: Quadrilateral $ASA'M$ is cyclic. Proof: Note that $\angle ASA'=180^\circ-\angle AVX=180^\circ-\angle AMA',$ and so $ASA'M$ is cyclic $\blacksquare$ Claim 4: Quadrilateral $SX'MX$ is cyclic. Proof: Note that $\angle X'XS=\angle AA'S=\angle AMS,$ and so $SX'MX$ is cyclic $\blacksquare$ Now, let $Q$ be on $BC$ such that $QX$ is perpendicular to $XM$. By symmetry, $QX'$ is perpendicular to $X'M$. Note that quadrilateral $QX'MX$ is cyclic, and so points $Q,S,X',M,X$ are all concyclic, therefore $\angle QSM=\angle QX'M=90^\circ,$ thus $\angle QSM=90^\circ=\angle ASM,$ which in turn gives that points $Q,A,S$ are collinear. To conclude, if $SX'$ intersects $(ABC)$ at $Z'$, then $\angle AXZ'=\angle ASZ=\angle ASX'=\angle AMQ=\angle AVX=\angle AZ'X,$ and so $AX=AZ',$ which implies that $Z \equiv Z',$ as desired.

Including $X'$ suggests that it's the reflection of $X$ over $\overline{BC}$ which turns out to be true. Anyway, config geo never dies I guess, here's a solution: Getting rid of $V$, $E$, $F$: These points are rather artificial. Notice that \[\angle MXE = \angle MFE = \frac{\overarc{$BE$}}{2}+\frac{\overarc{$AC$}}{2} = \frac{\overarc{$CE$}}{2}+\frac{\overarc{$AC$}}{2} = \angle AVE = \angle MVE\]Hence points $V$ and $X$ are symmetric wrt line $\overline{ME}$. Now $\angle AMC = \angle XMC$, so $ABXC$ is harmonic (this is a well-known property) and we can forget about points $V, E, F$ as we've categorized the points $X$ and $X'$ (as $V$ and $X$ are symmetric wrt $\overline{ME}$ it follows that $MX' = MV = MX\Longrightarrow X'$ is the reflection of $X$ across $\overline{BC}$ as conjectured above). Characterizing S: Let $B', C'$ be the feet of the altitudes from $B, C$ in $\triangle ABC$ and $Y = \overline{B'C'}\cap\overline{BC}$. Then notice that: \[(B,C;A',\overline{AS}\cap\overline{BC}) \overset{S}{=} (B,C;X,A) = -1 = (B,C;A',Y) \Longrightarrow Y\in\overline{AS}\]It's well-known that this implies that $S$ lies on the line $\overline{HM}$ where $H$ is the orthocenter of $\triangle ABC$. Moreover, $S$ is the Miquel point of the cyclic quad $BCB'C'$ and $\angle MSY = 90^{\circ} = \angle MX'Y = \angle MXY$ ($H$ is the orthocenter of $\triangle AMY$ as $AH\perp MY$ and $MH\perp AY$), so $XMX'SY$ is cyclic. Now if $\angle AMC = \varphi$, then \begin{align*} \angle X'SX &= 180^{\circ} - \angle X'MX \\ &= 180^{\circ} - 2\varphi \\ &= 2\angle AMB - 180^{\circ} \\ &= 2\angle ACX - 180^{\circ} \\ &= 180^{\circ}-2\angle AZX \\ &= \angle ZAX \\ &= \angle ZSX \end{align*}So $\angle X'SX = \angle ZSX$ and points $S, X', Z$ are collinear, as desired.