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How many triples $(x,y) \in \mathbb{N^+}^2$ and $n \gt 1$ are there such that $x^n - y^n = 2^{100}$

I dont know how to start. Any hint will be helpful.

I know the identity $x^n-y^n = (x-y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2}+y^{n-1})$.
I think from here we need some combinatorics to get the rest of answer.

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    Notice the equation in the title is not the same as the one in the body of your Question. Please fix this and add any observations you made about existence of solutions or useful identities. – 2017-02-02
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    Title differs from question. – 2017-02-02
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    $x^n + y^n = 2^{100}$ or $x^n - y^n = 2^{100}$? The body of your problem does not agree with your title. – 2017-02-02
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    opps ... fixing that .. :) – 2017-02-02
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    for $n>2$ you can invoke Fermat's last theorem and, at least remove from consideration many values of $n$ – 2017-02-02
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    @DougM Not that many; just those that divide $100$. – 2017-02-02
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    I think this falls to the lifting the exponent lemma. – 2017-02-02
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    I didn't want to plagiarise, so [here](http://www.qbyte.org/puzzles/p105s.html) is your answer! Hope it helps. – 2017-02-02
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    @Rohan Thanks :) – 2017-02-02
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    @RezwanArefin No problem. – 2017-02-02
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    @Rohan I didn't get this why they conclude $, (a, b) = (1, 99), (2, 98), ... , (49, 51)$. But why $(50,50)$ is not there? And how they got this :| – 2017-02-02
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    We have that if $(x-y)=(x+y)=2^{50} $, then we will get $x=2^{50} $ and $y=0$. But both should be positive integers which is not satisfied by $y $. Hence, $(50,50) $ is not permitted. – 2017-02-02
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    @Rohan I don't understand why that is plagiarism. Plagiarism, according to the Oxford Dictionary, is defined as " The practice of taking someone else's work or ideas and passing them off as one's own." If you cite the source, it is hardly plagiarism. Of course, I think that the words were not your own, you should make it a community wiki answer. – 2017-02-02

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