540 = 5 * 2^2 * 3^3.
The 5 can be in either of the integers, or both: that's 3 possibilities.
The 2^2 can be in either or both of the integers, but if it's in only one of them, the other integer could have or not have a factor of 2. So the next factor by which to multiply the possible arrangements is (2*2 + 1), the solitary 1 being the case of both having the 2^2 as a factor, and the 2*2 for the choice of which one has the 2^2, and whether the other integer is or is not even.
So far we have 3*(2*2+1).
The 3^3 can be in either or both the integers. If it's in only one or the other, the one it's not in could be a multiple of 9, or of just 3, or no factor of 3 at all. So the next factor by which we multiply the possibilities is (2*3+1).
That makes 3*(2*2+1)*(2*3+1) = 105.
However, that considers the order of the integers: 1*540 and 540 * 1 are considered separate instances.
The only pair of integers that has not been counted twice is 540*540.
So 105 = 104 + 1, but the 104 must be halved, as we don't want order to matter.
The answer is 52 + 1 = 53.