Solution:

Answer: 49. Break all possible values of $n$ into the four cases: $n = 2$, $n = 4$, $n > 4$ and $n$ odd. By Fermat's theorem, no solutions exist for the $n = 4$ case because we may write $y^{4} + (2^{25})^{4} = x^{4}$.

We show that for $n$ odd, no solutions exist to the more general equation $x^{n} - y^{n} = 2^{k}$ where $k$ is a positive integer. Assume otherwise for contradiction's sake, and suppose on the grounds of well ordering that $k$ is the least exponent for which a solution exists. Clearly $x$ and $y$ must both be even or both odd. If both are odd, we have $(x - y)(x^{n-1} + \ldots + y^{n-1})$. The right factor of this expression contains an odd number of odd terms whose sum is an odd number greater than 1, impossible. Similarly if $x$ and $y$ are even, write $x = 2u$ and $y = 2v$. The equation becomes $u^{n} - v^{n} = 2^{k-n}$. If $k-n$ is greater than 0, then our choice $k$ could not have been minimal. Otherwise, $k-n = 0$, so that two consecutive positive integers are perfect $n$th powers, which is also absurd.

For the case that $n$ is even and greater than 4, consider the same generalization and hypotheses. Writing $n = 2m$, we find $(x^{m} - y^{m})(x^{m} + y^{m}) = 2^{k}$. Then $x^{m} - y^{m} = 2^{a} < 2^{k}$. By our previous work, we see that $m$ cannot be an odd integer greater than 1. But then $m$ must also be even, contrary to the minimality of $k$.

Finally, for $n = 2$ we get $x^{2} - y^{2} = 2^{100}$. Factoring the left hand side gives $x - y = 2^{a}$ and $x + y = 2^{b}$, where implicit is $a < b$. Solving, we get $x = 2^{b-1} + 2^{a-1}$ and $y = 2^{b-1} - 2^{a-1}$, for a total of 49 solutions. Namely, those corresponding to $(a, b) = (1, 99), (2, 98), \cdots, (49, 51)$.