let N = 2^n. Then 2N = 2*2^n = 2^(n+1)

then the problem is to find x and n (and by extension N) where:

x^n + N + 1 [call this A] divides x^(n+1) + 2N + 1 [call this B]

First let's get rid of some obvious candidates. What if x=1? Then we've got 1 + N + 1 dividing 1 + 2N + 1

Or N+2 dividing 2N + 2 = N+2 + N. Since N+2 divides N+2 and N+2 > N and hence never divides it, there are no solutions when x=1

Now suppose x = 2. Then x^n = 2^n = N and x^n+1 = 2*x^n = 2N, and we have N + N + 1 dividing 2N + 2N + 1 or 2N + 1 dividing 4N + 1. But 4N+1 = 2(2N+1) - 1 and so it can't possibly have 2N+1 as a divisor if N>0. No x=2 is never a solution.

Now, if A divides B then it must also divide kB where k is an integer. In particular, it must divide (N+1)B.

Further, if A divides B (or kB) then it must also divide kB-mA where m is an integer. In particular, the divisor must hold when m = (2N+1)

A, then, must divde (N+1)B - (2N+1)A = (N+1)(x^(n+1) + 2N + 1) - (2N+1)(x^n + N + 1)

= (N+1)x^(n+1) - (2N+1)x^n + (2N+1)(N+1) - (2N+1)(N+1)

= x^n[(N+1)x - (2N+1)] (Call the term in [] C)

It's a fair question to ask whether the subtraction has resulted in a number < 0, which may possibly cause trouble down the road. But we're now equipped to handle this case. Consider when (N+1)x = (2N+1), for this is the point below which C is negative. A bit of algebra gives x = 2N+1 / N+1 = (N + N+1)/(N+1) = 1 + N/N+1. Since N/N+1 < 1, the point where C=0 occurs when x < 2. Well, we just showed there are no solutions when x <= 2 so we can safely assume C > 0.

Now, suppose A and x^n have a common factor f. Then since A = x^n + (N+1), f divides N+1 as well. And since f divides N+1 it does NOT divide N.

Now consider terms in B: by assumption f divides x^n so it also divides x^(n+1). f divides (N+1), and B = x^(n+1) + (N+1) + N. Since B is then NOT divisible by f (it's one less than a multiple of f) but A IS divisible by f, A cannot divide B. So, if A divides B, then x^n and A can share no common factor. A must divide Cx^n but it has no common factors with x^n and so: A must divide C.

That's an interesting result, since A has a term of x^n but C is linear in x -- yet for A to divide C it must be true that A <= C. More specifically, C > 0 and C = kA for some positive integer k.

For any specific n, compare A-C for consecutive values of x. That is, compare (x^n + N + 1 - (N+1)x + (2N+1)) with ((x+1)^n + N + 1 - (N+1)(x+1) + (2N+1)). The difference is (x+1)^n - x^n - (N+1)(x+1) + (N+1)x = (x+1)^n - x^n - (N+1). When is this difference positive? Well, if n=1, then this is a constant expression whose value is -2. It's NEVER positive. When n = 2, we instead have (x+1)^2 - x^2 - 5 = 2x + 1 - 5 = 2x - 4 which is positive when x >= 2. And as n gets larger, we approximately have x^(n-1) = 2^n for the zero, or x ~ 2*[2^1/(n-1)] which is less than 3. since the second factor is less than 1.5. The terms on the LHS that were omitted that are lower order polynomial terms in x are larger than the constant 1 that was omitted on the RHS. Since the LHS is actually bigger than the approximation, the real boundary is smaller than the approximate one here; the inequality holds.

OK, well, if A-C is positive, then A > C and hence A can't divide C. So if n >= 2 the only way A < C is if x <= 2. But we already showed there are no solutions when x <= 2 so if n >= 2 there are no solutions at all!

We now just have to tie up the case of n=1 (N=2^1=2). For this case, C = 3x - 5 and A = x+3 so we have C = kA or 3x-5= k(x+3). Now, if k = 3 there is no solution, and if k > 3 the only solutions are when x < 0. So we need only consider k=1 and k=2. For k=1, we have 3x-5=x+3 and x = 4. For k=2, we have 3x-5=2x+6 and x=11. So (4,1) and (11,1) are solutions, and these are the only solutions when n=1. Further, since there are no solutions at all when n > 1 these are the only solutions period.