how to solve the following mordell equation:$ y^2 = x^3 - 3$

Solution 1:

Yes, I think this is the best way. Keith Conrad has answered this with many examples in his article Examples of Mordell's equation; e.g., compare with Theorem $2.2$ and $2.3$, which is very similar, for $y^2=x^3-5$ and $y^2=x^3-6$.

Solution 2:

Your solution is flawed, so here's a complete solution: As you said, $x$ is odd, because if $x$ were even, then $y^2\equiv -3\pmod{8}$, which is not a quadratic residue. Then $y=2k$ is even.

$4(k^2+1)=(x+1)\left(x^2-x+1\right)$.

$x^2-x+1$ is always odd. Let $p$ be a prime divisor of $x^2-x+1$. Then $p\equiv 1\pmod{4}$, because if $p\equiv 3\pmod{4}$, then $k^2\equiv -1\pmod{p}$, contradiction (by Quadratic Reciprocity). Therefore $x^2-x+1\equiv 1\pmod{4}$, so $4\mid x(x-1)$, so $4\mid x-1$, i.e. $x\equiv 1\pmod{4}$. But then $x+1\equiv 2\pmod{4}$, contradiction, because this implies $4\nmid (x+1)\left(x^2-x+1\right)$ (while $4$ divides the LHS). No solutions exist.