• It actually has two solutions, the other one being $\angle PXQ = 180 - \angle PYQ$. You can try verifying this case, but the question statement is not true in this case. So the question is partially false.
• It is recommended that you read EGMO (Euclidean Geometry for Mathematical Olympiads) Chapter 1

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.