<s>[INST] How many triples $(x,y,n)$ are there such that $x^n - y^n = 2^{100}$ 
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. 
 [/INST] Since the question is being asked rather often and the solution in the comments seems to have a flaw I propose the following answer.

We claim: the equation
$$
a^n-b^n=2^k,\tag1
$$
where $a,b,n,k$ are positive integers ($n>1$), has exactly $\left\lfloor\frac{k-1}2\right\rfloor$ solutions $(a,b)$ for $n=2$ and no solutions for $n>2$.
First of all we note that from
$$
a^n-b^n=(a-b)\sum_{i=0}^{n-1}a^{n-1-i}b^i=2^k,\tag2
$$
it immediately follows that the solution can exist for $n>1$ only if $k>0$.
Now let consider the problem case by case.
For $n=2$ from
$$
a-b=2^r,\quad a+b=2^{k-r}
$$
the only solutions are:
$$a=2^{k-r-1}+2^{r-1};\quad b=2^{k-r-1}-2^{r-1},\quad 0<r<\frac{k}2.\tag3$$
For $n=4$ we would have the equation:
$$
(a^2-b^2)(a^2+b^2)=2^k.
$$
This implies
$$a^2-b^2=2^m\implies a=2^{m-r-1}+2^{r-1};\quad b=2^{m-r-1}-2^{r-1}$$
for some $m$ and $r$.
But the equation
$$
2^{2(m-r-1)}+2^{2(r-1)}=\frac{a^2+b^2}2=2^{k-m-1}
$$
cannot hold in view of $r\ne\frac m2$ (cf. (3)).
For odd $n>1$ we note that equation (2) can hold only if both $a$ and $b$ are even: $a=2A$, $b=2B$. Assume now that equation (1) has an integer solution $(a,b)$ for some $k$. Then there must exist some minimal positive $K$ for which the equation holds. However from
$$
(2A)^n-(2B)^n=2^K\implies A^n-B^n=2^{K-n}
$$
it follows that such minimal $k$ does not exist (since $n<K$).
In general case an arbitrary $n>2$ can be represented as $2^p q$, where $q$ is an odd number. If $p>0$ repeating the process:
$$
2^k=a^n-b^n=(a^{n/2}-b^{n/2})(a^{n/2}+b^{n/2})\implies a^{n/2}-b^{n/2}=2^m;\ (
m>0)$$
one reduces the problem either to $n=4$ (if $q=1$) or $n$ odd (if $q\ne1$). The obtained contradiction finalizes the proof.
</s>