Solution:
Let $D$ be the foot of the altitude from $A$. Let $H$ be the orthocenter of triangle $ABC$. Let $M$ be the midpoint of $AH$. Let $I$ be the incenter of triangle $ABC$. Let $\omega$ be the incircle of triangle $ABC$. Let $\gamma$ be the circumcircle of $AEF$. Let $\eta$ be the nine-point circle of triangle $ABC$.

We first dispense with the case in which $AB = AC$. Since $M$ is the circumcenter of triangle $AEF$, if the circumcenter of triangle $AEF$ lies on the incircle of triangle $ABC$, then $\omega$ and $\eta$ intersect in two points: $M$ and $D$. Since $\omega$ and $\eta$ are tangent by Feuerbach's Theorem, it follows that $\omega$ and $\eta$ are coincident, in which case triangle $ABC$ is equilateral.

If the incenter of triangle $ABC$ lies on the circumcircle of triangle $AEF$, then $A$, $I$, and $H$ lie on $\gamma$. Since no circle may intersect a line in three distinct points and $A$ is distinct from $I$ and $H$, it follows that $I$ and $H$ are coincident, in which case triangle $ABC$ is equilateral. Since it is clear that if $ABC$ is equilateral, $M$ lies on $\omega$ and $I$ lies on $\gamma$, this completes the proof in the case that $AB = AC$.

Throughout the remainder of this solution, we assume that $AB \neq AC$. Let $\Omega$ be the circumcircle of triangle $ABC$, and let $\alpha$ be the $A$-mixtilinear incircle. Let $\alpha$ be tangent to $AB$ at $P$, to $AC$ at $Q$, and to $\Omega$ at $R$. Since $AB \neq AC$, lines $PQ$ and $BC$ are not parallel. Let their intersection be $Z$, and let the bisectors of $\angle A$, $\angle B$, and $\angle C$ intersect $\Omega$ at $T$, $U$, and $V$, respectively. Let the point diametrically opposite $T$ on $\Omega$ be $S$. Let lines $RS$ and $BC$ intersect at $Y$. We introduce a lemma here, namely that $I$ lies on $PQ$ and that $Y$ and $Z$ are harmonic conjugates with respect to $B$ and $C$.

Let lines $AR$ and $PQ$ intersect at $K$. Since $PA$ and $QA$ are tangent to $\alpha$ at $P$ and $Q$, respectively, $RK$ is the symmedian from $R$ in triangle $PQR$. By Pascal's Theorem applied to cyclic hexagon $ACVRUB$, points $Q$, $I$, and $P$ are collinear. Since $AP = AQ$ and $AI$ bisects $\angle PAQ$, $I$ is the midpoint of $PQ$. Hence $RI$ is the median from $R$ in triangle $PQR$. It follows that $\angle ICQ = \angle PRA = \angle IRQ$ and $\angle IBP = \angle QRA = \angle IRP$. Hence quadrilaterals $RBPI$ and $RCQI$ are cyclic, so $RI$ is the bisector of $\angle BRC$. It follows also that $\angle BIP = \angle BCI$, so line $PQ$ is tangent to the circumcircle of triangle $BIC$ at $I$. Since $T$ is the circumcenter of triangle $BIC$ and $S$ is diametrically opposite $T$ on $\Omega$, $BS$ and $CS$ are tangent to the circumcircle of triangle $BIC$ at $B$ and $C$, respectively. It follows that $IS$ is the symmedian from $I$ in triangle $BIC$; since $RI$ is the bisector of $\angle BRC$, $R$, $I$, and $S$ are collinear. Hence $RS$ is the polar of $Z$ with respect to the circumcircle of triangle $BIC$, so $Y$ and $Z$ are harmonic conjugates with respect to $B$ and $C$, as desired.

We now take up the main problem. We first prove that $I$ lies on $\gamma$ if $M$ lies on $\omega$. If $M$ lies on $\omega$, then $M$ is the Feuerbach point. Let $X$ be the center of the homothety of positive magnitude mapping $\omega$ to $\Omega$. Since the center of the homothety of positive magnitude $\omega$ to $\eta$ is $M$ and the center of the homothety of positive magnitude mapping $\eta$ to $\Omega$ is $H$, it follows by Monge's Circle Theorem that $X$ lies on line $MH$. Note that $R$ is the center of the homothety of positive magnitude mapping $\alpha$ to $\Omega$. Since the center of homothety of positive magnitude mapping $\alpha$ to $\omega$ is $A$, it follows by the same theorem that $R$ lies on line $AX$, so points $A$, $M$, $H$, and $R$ are collinear.

Note that triangle $BRC$ is the reflection of triangle $BHC$ about line $BC$. Since line $RY$ bisects angle $BRC$, it follows that line $HY$ bisects $\angle BHC$. Let the line through $H$ parallel to $PQ$ intersect $BC$ at $Z'$. Since $\angle HBA = \angle HCA$, the internal bisectors of $\angle BAC$ and $\angle BHC$ are parallel. Since $PQ$ is perpendicular to $AI$ and $BZ'$ is parallel to $PQ$, it follows that $BZ'$ is the external bisector of $\angle BHC$. Hence $Y$ and $Z'$ are harmonic conjugates with respect to $B$ and $C$. It follows that $Z$ and $Z'$ are coincident, and, therefore, that $H$ and $K$ are coincident. We may conclude that $\angle AIH$ is right, which implies the desired result because $\gamma$ is the circle on diameter $AH$. (Do think carefully about why this argument fails if $AB = AC$.)

Conversely, if $I$ lies on $\gamma$, then $\angle AIH$ is right, so points $P$, $H$, $I$, and $Q$ are collinear. Since $PQ$ is perpendicular to $AI$ and the internal bisectors of $\angle BAC$ and $\angle BHC$ are parallel, it follows that $PQ$ is the external bisector of $\angle BHC$. Let the internal bisector of $\angle BHC$ intersect $BC$ at $Y'$. Then $Y'$ and $Z$ are harmonic conjugates with respect to $B$ and $C$, so $Y$ and $Y'$ are coincident. Let $R'$ be the reflection of $H$ about line $BC$. Clearly $R'Y$ is the bisector of $\angle BR'C$, so $R'$, $Y$, and $S$ are collinear. Since $R$, $Y$, and $S$ are collinear and $R'$ also lies on $\Omega$, it follows that $R$ and $R'$ are coincident. Since $HR'$ is perpendicular to $BC$, it follows that points $A$, $H$, and $R$ are collinear.

Let $X'$ be the center of the homothety of positive magnitude mapping $\omega$ to $\Omega$. Since the center of the homothety of positive magnitude mapping $\omega$ to $\alpha$ is $A$, and the center of the homothety of positive magnitude mapping $\alpha$ to $\Omega$ is $R$, it follows by Monge's Circle Theorem that $X'$ lies on line $AR$. Note that $H$ is the center of the homothety of positive magnitude mapping $\Omega$ to $\eta$. Since the center of the homothety of positive magnitude mapping $\omega$ to $\eta$ is the Feuerbach point, it follows by the same theorem that the Feuerbach point lies on line $AR$ as well. Since $AR$ intersects $\eta$ at $M$ and $D$, the Feuerbach point may be either $M$ or $D$. However, $D$ may not be the Feuerbach point, else the altitude from $A$ and the bisector of $\angle A$ in triangle $ABC$ would be coincident, contradicting the assumption that $AB \neq AC$. We may conclude that $M$ is the Feuerbach point, so, in particular, $M$ lies on $\omega$, as desired. This completes the proof.