Integral solutions to $y^{2}=x^{3}-1$ How to prove that the only integral solutions to the equation $$y^{2}=x^{3}-1$$ is $x=1, y=0$. 
I rewrote the equation as $y^{2}+1=x^{3}$ and then we can factorize $y^{2}+1$ as $$y^{2}+1 = (y+i) \cdot (y-i)$$ in $\mathbb{Z}[i]$. Next i claim that the factor's $y+i$ and $y-i$ are co-prime. But i am not able to show this. Any help would be useful. Moreover, i would also like to see different proofs of this question.
Extending Consider the equation $$y^{a}=x^{b}-1$$ where $a,b \in \mathbb{Z}$ and $(a,b)=1$ and $a < b$. Then is there any result which states about the nature of the solution to this equation.

Is it really necessary to use complex numbers here?
Does this sound reasonable to you or is there a flaw?
According to the 7-period for n^2 and n^3-1 (mod 7) 
LHS=RHS requires both to be 0 mod(7), so that for some n and k 
y^2=n*7^2 must equal  k* 7^3 -1  and therefore 7^2(n-7k)=-1 , impossible, since 7^2>1. 
I note the table format doesn't come out right - you have to do it yourselves.
