Cubic diophantine equation

Perhaps others can find an elementary solution. This is an elliptic curve, of rank $1$ and trivial torsion subgroup. So the group of all rational points is generated by one single rational point, namely $P=(1,1)$. Hence, every rational point on the curve is of the form $nP$ for some $n\in\mathbb{Z}$. Here are some multiples: $$2P=(2 , -3 ),\ 3P=(13 , 47),\ 4P=\left(\frac{25}{36} , \frac{37}{216}\right),\ 5P=\left(\frac{685}{121} , -\frac{18157}{1331}\right),\ldots $$ Some cumbersome arguments using heights can show that, in fact, the point at infinity, together with $\pm P$, $\pm 2P$, and $\pm 3P$ are the only integral points on the curve (since this curve is of the form $y^2=f(x)$, if $Q=(x_0,y_0)$ is a point on the curve, then $-Q=(x_0,-y_0)$).