Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

Let $R_1,R_2$ denote the circumradius of $\triangle XPQ,\triangle YPQ,$. So we have 

$R_1 = R_2$

$\Rightarrow \frac{PQ}{sin\angle PXQ}=\frac{PQ}{sin\angle PYQ}$

$\Rightarrow \angle PXQ = \angle PYQ(*)$

Let $Y'$ be the reflection of Y across PQ and $J'$ be the incenter of $\triangle PY'Q$. So $\angle PY'Q =\angle PYQ =\angle PXQ $. So, $PXY'Q$ is cyclic. Let $K = (XPQ) \cap IX$. 

So we have, $\angle IPK = \angle IPQ + \angle QPK  = \angle IPX + \angle QXI = \angle IPX + \angle PXI = \angle KIP$. 

So, $PK = IK$, similarly $QK = IK$. So $K$ is the midpoint of arc $PQ$. Similarly, $Y'J'$ also passes through $K$.

So there is a circle centered at $K$ that goes through $P,Q,I,J'$. So PQIJ' is cyclic. Because of the fact that I,J,M are collinear, we have $\angle IMP = \angle J'MQ$. So we have $\triangle KMI \cong \triangle KMJ'$. So, $IM$ = $J'M$ $\Rightarrow \triangle IMP = \triangle J'MQ$ $\Rightarrow \angle MPI = \angle MQJ'$ $\Rightarrow \angle XPQ = \angle Y'QP$. So $XY'QP$ is a cyclic trapezoid. Hence, $XPYQ$ is a parallelogram.