I understand the rest of the solution from the tutorial, but I do not understand how to do this "The first step, as in the solution, is to find all divisors of all numbers in the range in mlogm time". I am just a beginner. What I can think of is that we use seive to find all prime factors for each number, but I do not understand how to go from here to finding all factors.

Yes, you are right! That makes absolute sense and is a much simpler way to solve this. Even in my original solution, I did not observe that (1,2) is the only solution, instead came up with a way to efficiently find all possible solutions using sorting and two pointers(based on the growth rate of exponents), Although I kept getting WA due to integer overflow(facepalm).

But I feel that without this observation we end up doing more computation than is required(even though asymptotically it is the same) so I just wanted to wrap my head around it.

For those who feel(like me) that the observations in problem D are hard to make analytically(and at the same time be sure that no edge case was missed), we can use high school calculus to solve this in a more structured and generalized format, which may be useful in some other questions-

After you get the equation a⋅2^b=b⋅2^a, rewrite it as (a.2^-a)=(b.2^-b). This holds when a=b. Furthermore, we ask, "For what distinct integral values of x is f(x)=x.2^-x equal to itself?" We can see this by making a rough plot of f(x). There is only one critical point(x=(1/ln2)~1.4), which is a maximum (we can easily observe this using just the first derivative). So the curve looks like a bell, strictly increasing before 1, peaking between 1 and 2, and strictly decreasing thereafter.

The equality cannot hold on either side of the peak since we have a strict ordering. Therefore the only possible equality is if f(1) matches with some x>=2. Now we can either find that this holds for x=2 by trial and error, or more rigorously...wait for it...binary search. Search on the strictly decreasing array for f(i) i>=2 for the value of f(1), to get x=2. Then we can count the occurrences using standard techniques.