Note that z is odd. Then z2 ≡ 1 (mod 4). This implies that
7x ≡ 1 (mod 4). Thus, x is even. Let x = 2k where k is a positive integer.
Then z2 − 72k = 23y. Then (z − 7k)(z + 7k) = 23y. Thus, z − 7k = 2u where u
is a non-negative integer. Then z + 7k = 23y−u. Thus, 2(7k) = 23y−u − 2u =
2u(23y−2u−1). It follows that u = 1. Then 7k = 23y−2−1. Thus, 23y−2−7k = 1.
Since k ≥ 1, we have 3y − 2 ≥ 3. By Proposition 2.1, we have k = 1. Then
23y−2 = 8. Then 3y − 2 = 3. Thus, y = 5
3 . This is a contradiction.