Aren't the powers too large?

This isn't a complete solution, and I am not sure if a simple solution can be written down. By several papers due to Darmon-Merel, Bruin, Kraus, and Bennet (see references below), the equation

$x^n + y^n = z^{100}$

has no non-trivial solutions for $n \geq 3$ . However, this is only relevant to the problem when $n$ is odd; the even case is not covered since $(-y)^n = y^n$ in that case. One can probably adapt the same methods however to deal with the negative sign in that situation.

If there is a particularly novel solution I would love to see it.

Anyway, if we can settle on the fact that there are no solutions when $n \geq 3$ , then the rest is easy. We write $x^2 - y^2 = 2^{100} = (x-y)(x+y).$ Then we must have $x - y = 2^k,$ $x+y = 2^{100-k}.$ Solving the linear system, we see that $x = 2^{100-k-1} + 2^{k-1},$ $y = 2^{100-k-1} - 2^{k-1}.$ From here we see that $k -1 < 100 - k - 1,$ which implies that $1 \leq k \leq 49$ . Thus there are $49$ choices.

References:

M.A. Bennet, The equation $x^{2n} + y^{2n} = z^5$ , J. Th. Nombres Bordeaux 18 (2006), 315-321.

N. Bruin, Chabauty methods using elliptic curves, J.reine angew. Math. 562 (2003), 27-49.

N.Bruin, On powers as sums of two cubes, ANTS IV, Leiden 2000, 169- 184, Lecture Notes in Comput.Sci. 1838, Springer 2000.

H.Darmon, L.Merel, Winding quotients and some variants of Fermat’s Last Theorem, J.reine angew.Math. 490 (1997), 81-100.

A.Kraus, Sur l’´equation a3 + b3 = cp, Experiment.Math.7 (1998), 1-13.

We show a stronger result:

The equation $x^n-y^n=2^k$ , $n\geq 2$ has a solution $(x,y,n,k)$ in naturals iff $n=2$ .

Proof: Let $g = gcd(x,y)$ . Then $g|2^{k}$ , so $g = 2^{l}$ , for some non-negative integer $l$ .

So take $x=gp , y=gq$ , to get $p^n-q^n =2^{k-l}$ . Note that $k \neq l$ ; if it was, then $p^n-q^n =1$ , which is clearly impossible for $n\geq 2$ . So $k > l$ .

Therefore , $p^n - q^n$ is even , so both $p , q$ are odd, so $2|(p-q)$ .

This means the exceptional case of the Zsigmondy Theorem must apply, so that $n=2$ and $p+q =2^{m}$ for some positive integer $m$ .

So now $2^{m}(p-q) = (p+q)(p-q) = p^2 - q^2 = 2^{k-l}$ , so $p- q = 2^{k-l-m}$ . But now, we will get $2p = 2^{m} + 2^{k-l-m}$ , but $p$ is odd, so this immediately means that $2^{k-l-m} =2$ . So $p-q =2$ .

Therefore, we have $p = 2^{m-1} + 1$ and $q= 2^{m-1} -1$ . This means $x= (2^l)(2^{m-1} + 1)$ and $y =(2^l)(2^{m-1} -1)$ . But $y \geq 1$ , so $m\geq 2$ .

$x^2-y^2 = 2^{k}$ gives $2l+m+1 =k$ .

Here $k=100$ , so $2l+m =99$ . We get our $49$ pairs $(l,m) =(0,99), (1,97),...., (48,3)$ , and the corresponding $(x,y)$ .

It is obvious that we can set n to be a prime p.Lets consider the general case when 100 is replaced by m and x and y are coprime ( if they had a common divisor, it would be a power of two.But the lhs is homogenous, so we return to an equation of the previous type).The LHS is factored as follows:

$(x-y)(x^{n-1}+...+y^{n-1})=2^m$

The first factor is obviously smaller than the second one, and both are powers of two, so the smaller divides the greater.But $x\equiv y\pmod{x-y}$ , so the second factor is congruent to $nx^{n-1}$ modulo $x-y=2^a$ .But we required x and y be coprime, so both are odd.That means from the last congruence that $2^a|n$ ,n is prime.

Case 1.n is odd Here a=0 and thus $x=y+1$ .We have the equation $(y+1)^n-y^n=2^m< (y+1)^n \leq 2^n$ , so n>m.But when expanding, the highest order term is $n.y^{n-1}$ , whis is less than $2^m$ only for y=1 by the previous argument.But this is impossible.

Case 2.n=2

This case easily leads to $x=2^{m-2}+1, y=2^{m-2}-1$ .So to solve $X^2-Y^2=2^{100}$ , we must have that $X=2^{100-m}.x,Y=2^{100-m}.y$ .By a simple counting argument we arrive at the answer 49.