Here is my question:
Find all solutions to $y^2=x^3+4$.
My attempt:
Rewrite the equation as $(y-2)(y+2)=x^3$. Notice that if $y$ is odd, then $(y-2,y+2)=1$. Hence they are both cubes, but no cubes differ by $4$. Therefore $y$ is even and $x$ is even too.
Write $y=2y'$, then $4y'^2=x^3+4$. Write $x=2x'$, and we finally deduce $$y'^2=2x'^3+1.$$
However, I cannot proceed after this step. Can anyone help? Thanks. The solution should be mainly done with elementary steps.
If you are looking for integer points, then the LMFDB curve 108.a2 (Cremona label 108a1) $\,y^2=x^3+4,\,$ has the information that $\,x=0, y=\pm2 \,$ are the only integer solutions. There is more information in the entry, such as the Mordell-Weil group structure is $\,\mathbb{Z}/3\mathbb{Z}.$
Continuing from your attempt, with $\,x=2x_1, y=2y_1\,$ and your last equation, gives $\,(y_1-1)(y_1+1) = 2x_1^3. \,$ If $\,y_1\,$ is even gives a contradiction, thus $\,y_1\,$ is odd. But now, if $\,x_1\,$ is odd gives a contradiction, thus $\,x_1\,$ is even. Thus, let $\,x_1 = 2x_2, y_1 = 2y_2 + 1.\,$ The equation is now $\,2y_2(2y_2+2) = 8y_2(y_2+1)/2 = 16x_2^3.\,$ This reduces to $\,y_2(y_2+1)/2 = 2x_2^3\,$ where the left side is a triangular number which is twice a cubic number (this is close to but not the same as a cubic triangular number). The obvious solutions are $\,x_2=0,y_2=0\,$ and $\,x_2=0,y_2=-1.$