Solution 1. Let $\Gamma$ be circle $ABC$ and $\omega$ be circle $AYZ$. Let $O, M$, and $S$ be the centres of $\Gamma, \Omega$, and $\omega$, respectively. Let $AK$ intersect $\Gamma$ again at $P$, and let the angle bisector of $\angle ZAY$ intersect $\omega$ again at $N$.

By power of a point from $K$ to $\Gamma$ and $\Omega$, we have that $KA \cdot KP = KB \cdot KC = KY \cdot KZ$, so $P$ also lies on $\omega$. The pairwise common chords of $\Gamma, \Omega$, and $\omega$ are then $AP \perp OS, BC \perp OM$, and $YZ \perp MS$, so we have that $\angle OMS = \angle CKY = \angle YKA = \angle MSO$. As $M$ lies on $\Gamma$ and $OM = OS$, $S$ also lies on $\Gamma$. Note that $N$ lies on $MS$ as $NY = NZ$, so
$$\angle PAN = \frac{1}{2} \angle PSN = \frac{1}{2} \angle PSM = \frac{1}{2} \angle PAM.$$
Thus, $AN$ bisects $\angle PAM$ in addition to $\angle ZAY$, which means that $\angle ZAK = \angle IAY$ as $K$ lies on $AP$ and $I$ lies on $AM$.

Solution 2. Define $M$ and $P$ as in Solution 1, and recall that $AYPZ$ is cyclic. Let $Q$ be the second intersection of the line parallel to $BC$ through $P$ with circle $ABC$ and let $J$ be the incentre of triangle $APQ$.

Since $PQ$ is parallel to $BC$ and $\angle BAP < \angle PAC$, the angle bisector of $\angle APQ$ is parallel to the angle bisector of $\angle AKC$. Hence, $PJ$ is parallel to $YZ$. As $M$ is the midpoint of $\overparen{PQ}$ on circle $APQ$, we have that $MP = MJ$. Then since segments $YZ$ and $PJ$ are parallel and have a common point $M$ on their perpendicular bisectors, $PJYZ$ is cyclic with $JY = PZ$. It follows that $J$ also lies on circle $AYPZ$ and that $\angle ZAP = \angle JAY = \angle IAY$.

Solution 3. As in the previous solutions, let $M$ be the centre of $\Omega$. Let $L$ be the intersection of $AM$ and $BC$, and let $L'$ be the reflection of $L$ over $YZ$. Let the circle $MYZ$ intersect $AM$ again at $T$.

Note that as $M$ is the midpoint of $\overparen{BC}$ on circle $ABC$ and $L$ is the foot of the bisector of $\angle BAC$, we have that $MA \cdot ML = MI^2 = MY^2$. It follows by power of a point that $MY$ is tangent to circle $ALY$, so $\angle LAY = \angle LYM$. Using directed angles, we then have that
$$\Varangle AYT = \Varangle MTY - \Varangle MAY = \Varangle MZY - \Varangle LYM = \Varangle ZYM - \Varangle LYM = \Varangle ZYL = \Varangle L'YZ,$$
where we use the fact that $MY = MZ$ and that $L$ and $L'$ are symmetric about $YZ$. Thus, $YT$ and $YL'$ are isogonal in $\angle AYZ$. Analogously, we have that $ZT$ and $ZL'$ are isogonal in $\angle YZA$. This means that $T$ and $L'$ are isogonal conjugates in triangle $AYZ$, which allows us to conclude that $\angle ZAK = \angle IAY$ since $L'$ lies on $AK$ and $T$ lies on $AI$.