Integral solutions to $y^{2}=x^{3}-1$

Alon Amit's answer is not right, I believe.

$2$ is not prime in the Gaussian integers! $(1+i)$ divides $2$.

The argument upto the fact that the prime $p$ divides $y+i$, $y-i$ and $2i$ is correct.

Since $p|2i$ we consider $p = 1+i$. Now any multiple of $1+i$ is of the form $2x + i2y$ or $2x+1 + i(2y+1)$. Since $y$ is even, $y+i$ cannot be divisible by $1+i$ and thus $y+i$ and $y-i$ are co-prime.

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.

If $x$ is even then $x^3$ is divisible by 4, so $x^3-1 \equiv 3 \pmod 4$ and this cannot be a square. Thus $x$ is odd and $y$ is even.

Now, if a prime $p$ divides both $y+i$ and $y-i$ then it divides their difference $2i$. Thus $p=2$ (up to units), but then $p$ divides $y$. That's impossible since it divides $y+i$ as well.

EDIT: I was too hasty in writing this, as pointed out by Moron. The prime 2 ramifies in $\mathbb{Z}[i]$, and is (up to units) the square of $1+i$. Sorry for the confusion.