The first one looks like case-work to me.
Since only positive integral solutions are asked for, the pairs (0,3) and (0,-3) are not counted.
We can rewrite the equation as

x=y gives no integer solutions.
Since the second term is even and is atleast 14 (since x and y are atleast 1 and x, y are unequal), (x-y)2 must be odd and less than or equal to 9
Hence the only possibilites are


Case 2 gets straightaway ruled out as y must be divisible by 3 and the least such +ve integer is 3 which means x = 0 which is not possible as x is also positive.
Case 1: y can be 1, 2 or 3.
y =1, x =2 is the only integral solution obtained
Hence (2,1) is the only solution among positive integers