Is it possible for $x^3-3$ to be a perfect square? 
Is it possible for $x^3-3$ to be a perfect square (or the negative of a perfect square), where $x$ is an integer?

I tried using modular arithmetic to find a contradiction, but couldn't find any. How can we find a contradiction?

The ring of Eisenstein integers is a Euclidean domain, hence a UFD. It follows that, in order to find the integer solutions of $(y+\sqrt{-3})(y-\sqrt{-3})=x^3$, it is enough to study the structure of cubes in $\mathbb{Z}[\omega]$ where $\omega=\frac{1+\sqrt{-3}}{2}$. Since
$$\begin{eqnarray*} (a+b\omega)^3 &=& (a^3-b^3-3ab^2)+3ab(a+b)\omega\\
&=&(a^3-b^3-3ab^2)+3ab(a+b)\frac{1+\sqrt{-3}}{2} \end{eqnarray*}$$
and $3ab(a+b)=\pm 2$ has no integer solutions, the Mordell equation
$$ y^2+3 = x^3$$
has no integer solutions, too.
