Given m and n are integers. They are cardinalities of two sets.
The total number of sub sets of first set: 2^m (I hope the null set is a subset of any set.). total number of subsets of second set: 2^n.
Hence, 2^m = 2^n + 60
The above equation is not satisfied by m =0 or 1, or n = 0 or 1.
So m and n are >= 2. Hence, 2^(m-2) and 2^(n-2) are also integers.
Now divide the above equation by 4. we get:
2^(m-2) = 2^(n-2) + 15
LHS is even as it is a power of 2. On RHS 15 is an odd number. So only way this equation can be solved is if 2^(n-2) is an odd number. The only possibility is n-2 = 0 for which it is equal to 1 and becomes an odd number. Hence,
n = 2. So m -2 = 4 and so m = 6
if you want a simple solution,
2^m = 2^n + 60
try n =0, 1, 2, 3,.. try to get RHS as a power of 2.
for n =2, RHS = 64 , it is a power of 2. and 2^6. Hence, m = 6.