In my last comment I looked at triplets of the exponents.  But actually the exponent on 2 is so large compared to 11 and 13 it is better to look at pairs of exponents for 11 and 13.

I will first illustrate with an example of 2024^3

2024^3 = 2^9 * 11^3 ^ 13^3

I can make a total of 16 distinct ordered pairs for the exponents of 11 and 13: (0,0) to (3,3).  Sort them in order of their sums and we will have 7 total classes:

(0,0)

(0,1), (1,0)

(0,2), (1,1), (2,0)

(0,3), (1,2), (2,1), (3,0)

(1,3), (2,2), (3,1)

(2,3), (3,2)

(3,3)

Next I want to add the exponent for 2, and I will do so by putting the larger exponents for 2 with the smaller exponents for 11 and 13.  Then the list looks like

Then these sets of exponents now form a set of 16 factors of 2024^3 such that no one of them is a divisor of another.

It is impossible to make this set larger.  Every possible pair of exponents for 11 and 13 is already present If I were to add any other factor, it must have a matching set of exponents for 11 and 13 which means that pair will be multiple of the one or the other by a factor of 2.

So then the answer for 2024^3 is 16, which is (3+1)^2

So now onto the problem as stated. 

2024^2024 = 2^6072 * 11^2024 * 13^2024

Identical setup, just larger.  Then the largest possible value of |S| for the problem as given is (2024+1)^2 = 4100625.