Find all integer solutions to $y^3 = x^6 + 6x^3 + 13$. Find all ordered pairs of integers $(x, y)$ that satisfy
$$y^3 = x^6 + 6x^3 + 13.$$
I've found the solutions $(-1, 2)$ and $(2, 5)$. I believe that these are all the integer solutions, but I don't know how to prove it. Could someone please help?

As Peter said in the comments, this is a special case of the Mordell equation $y^3=z^2+4$ but we can use the fact that $y$ is very nearly $x^2$ to obtain an elementary proof. 
$$\frac y{x^2}=\left(1+\frac 6{x^3}+\frac{13}{x^6}\right)^{1/3}$$
$$1-\frac 3{|x|^3}<\frac y{x^2}<1+\frac 3{|x|^3}$$
$$|y-x^2|<\frac 3{|x|}$$
And since $y-x^2$ is an integer, it is either 0 which implies $6x^3+13=0$ which is impossible or $\frac 3{|x|}>1$ which wields 5 cases that can be tested by hand. 
