I noticed while working on something else that 255 is 15*17, and 65535 is 255*257. In other words, it sounds like:

    2^(2^n)-1 * 2^(2^n)+1 = 2^(2^(n+1)) - 1

Testing some numbers, it looks like this works:

n	2^(2^n)-1	2^(2^n)+1	2^(2*(n+1)) - 1
0	1	3	3
1	3	5	15
2	15	17	255
3	255	257	65535
4	65535	65537	4294967295

And in fact we can prove that it holds for all n:

    2^(2^n)-1 * 2^(2^n)+1
       = 2^(2^n)*2^(2^n) + 2^(2^n) - 2^(2^n) - 1
       = 2^(2^n)*2^(2^n) - 1
       = 2^(2^n + 2^n) - 1
       = 2^(2*2^n) - 1
       = 2^(2^(n+1)) - 1

If 255 is 15*17 and 15 is 3*5, however, then as long as the numbers 3, 5, 17, 257, etc. are prime we can build up prime factorizations. So 255 would factor into 3*5*17 and 65535 would factor into 3*5*17*257. This suggests that if you have a number in the form 2^(2^n)-1 then its prime factorization is the product of 2^(2^i)+1 from i=0 to i=n-1:

n	2^(2^n)-1	prime factorization
1	3	3
2	15	3, 5
3	255	3, 5, 17
4	65535	3, 5, 17, 257
5	4294967295	3, 5, 17, 257, 65537

Neat!

But then I thought to try one more, and was very surprised:

n	2^(2^n)-1	prime factorization
6	18446744073709551615	3, 5, 17, 257, 641, 65537, 6700417

Why did our nice pattern break? It looks like 2^(2^5)+1 (or 4294967297) is 641*6700417. So not all numbers in the form 2^(2^n)+1 are prime, only the first five. The sequence is the Fermat numbers, integer sequence A000215. Such are the dangers of extrapolation.