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.

I think I have found a proof for the statement made in the top answer:

Setting $y+i$ to be a perfect cube (upto units), i.e. $y+i = (xi)^3$,
easily gives us $y=0$ and so $x=1, y=0$ is the only solution.

Let $y + i = (a + bi)^3 = (a^3 - 3ab^2) + i(3a^2b - b^3),$ then by comparison of coefficients, $1 = 3a^2b - b^3 = b(3a^2 - b^2)$, and therefore $b = \pm 1$.

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*Case 1: $b = 1\colon$ $3a^2 = 2 \Rightarrow a = \sqrt{2/3} \not\in \mathbb Z.$

*Case 2: $b = -1\colon$ $3a^2 = 0 \Rightarrow a = 0 \Rightarrow y+i = (-i)^3 = i \Rightarrow y = 0 \Rightarrow x = 1.$
