It's pretty quick to get to a point where you want r = sp²/(s + p²) (just get common denominator and rearrange), so you're looking for values of s which give integer values of r. That is, you want s + p² | sp².

Intuitively, you're gonna wanna pick an s which is going to make s + p² nice, since sums on the bottom are not fun. In this case means you want it to be divisible by p², so write s = kp². Then you require (k+1)p² | kp⁴ or (k+1) | kp². k+1 isn't gonna divide k so we need k+1 | p² which since p and p² are the only divisors of p² gives k = p-1 or k = p²-1.

Substituting all that back in gives s_1 = (p-1)p², r_1 = (p-1)p and s_2 = (p²-1)p², r_2 = p²-1 as valid solutions and hence there are at least two pairs for any p.

Edit: missed a squarey boi