Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$ Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n&gt;1$


I tried assuming it was equal to $\frac p q $.

I get $2^nq^2-q^2 = p^2 $ 

But I don't see where to go from there.
For $n\geq2$, $2^n-1\equiv 3\pmod 4$, unlike a perfect square.
It is clearly equivalent to prove that the diophantine equation $$2^n=x^2+1$$ has no integer solution for $n\gt 1$ Notice that $x$ must be odd so $x=2m+1$ hence
$$2^n=(4m^2+4m+1)+1=4m^2+4m+2\iff2^{n-1}=2m^2+2m+1$$ This is absurde.