First note that y = 0 gives no solutions, and that any other y solution has a matching solution for -y. So just find the solutions where y > 0 and then add their negative alternates.

Rewrite as 5*3^x = (y+2)(y-2)

y+2 and y-2 can't both be multiples of 3 (they differ by 4) so all of the factors of 3 on the left must be in just one of the terms on the right. The factor of 5 must be the other term (or else the two terms would differ by more than 4). If the larger term is 5 we have 5*1  and (x,y) = (0,1); if the smaller term is 5 we have 5*9 and (x,y) = (2,7).

The complete list of solutions, then, is: (0,-1), (0,1), (2, -7), (2,7)