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.

Let $\alpha\in\mathbb{Z}[i]$ be a divisor of $y+i$ and $y-i$. Then $\alpha|2=i(y-i-(y+i))$ and $\alpha|(y-i)(y+i)=x^3$. Since $x$ is odd then $x^3$ is odd and therefore, by Bezout there exist $A,B\in\mathbb{Z}$ such that $Ax^3+2B=1$ and therefore $\alpha|1$ implying $\alpha\in\mathbb{Z}[i]^{\times}$. We conclude that $y+i$ and $y-i$ are coprime.
