Maybe start by finding all pairs of integer factors of 102.

$\pm 1\ and\ \pm 102,\ \pm 2\ and\ \pm 51,\ \pm 3\ and\ \pm 34,\ or\ \pm 6\ and\ \pm 17.$

As you saw $x^2 - y^2 = 102 \implies (x - y)(x + y) = 102.$

$x\ and\ y\ are\ integers \implies (x - y) = m\ and\ (x + y) = n\ are\ integers.$

$x - y = m\ and\ x + y = n \implies x - y + x + y = m + n \implies x = \dfrac{m + n}{2} \implies |x| = \bigg |\dfrac{m + n}{2} \bigg|.$

$So\ x \in \mathbb Z\ only\ if\ |m + n|\ is\ even.$

$But\ mn = 102 \implies Signum(m) = Signum(n) \implies |m + n| = |m| + |n|.$

$So\ x \in \mathbb Z\ only\ if\ |m| + |n|\ is\ even.$

$|1| + |102| = 103,\ not\ even.$

$|2| + |51| = 53,\ not\ even.$

$|3| + |34| = 37,\ not\ even.$

$|6| + |17| = 23,\ not\ even.$

$So\ x \not \in \mathbb Z.$

$\therefore y \not \in \mathbb Z.$