prove that $2^q-1=x^2-y^2$ has only one solution

Question

Prove that $2^q-1=x^2-y^2$ has only one solution for a given value of $q$ where q is a prime and x,y are positive numbers. The obvious solution is $y=2^{q-1}$.

Answer 1

$$2^{11}-1=56^2-33^2=1024^2-1023^2$$

Answer 2

Let $N$ be a positive integer, how many ways are there to write it in the form $(x-y)(x+y)$ where $x$ and $y$ are positive integers? we must clearly have $x>y$.

Let $z=x-y$,this is a positive integer, we get $N=z(z+2y)$.

Clearly the number of ways to write $N$ in this form when $N$ is odd is equal to $\lceil\tau(N)/2\rceil$ ( here $\tau(N)$ is the number of divisors of $N$).

Therefore the solution is unique if and only if $N$ is a prime.

It is known that $2^q-1$ is not always a prime. In fact such primes are called mersenne primes and it is not even known if there are infinitely many.

Answer 3

I cannot see that there is only one solution. Already for $q=2$ and $q=3$ we have $2^2-1=2^2-1^2$ and $2^3-1=4^2-3^2$.

Edit: Let $q=37$. Then $2^q-1=137438953471\cdot 1=223\cdot 616318177$. This gives two different ways to write it as $(x+y)(x-y)$. So the solution is not unique for given $q=37$.