With the transformation $X = -n/x$ and $Y= ny/x$, the curve becomes isomorphic to the Weierstrass model $$ E_n\colon \ \ Y^2 - X\ Y - n\ X = X^3.$$ The points in question are exactly the integral points in $E_n(\mathbb{Q})$ such that $X$ divides $n$. I do not see why the number of these points should be bounded independently of $n$; so my guess is that there is no bound and that it is going to be difficult to show this.
The curve $E_n$ has always two rational 3-torsion points $(0,0)$ and $(0,n)$. Unless $n$ is of the form $k\cdot (\tfrac{k-1}{2})^2$ for some integer $k\not\equiv 2\pmod{4}$, these are all the torsion points in $E_n(\mathbb{Q})$, otherwise there are 6 torsion points defined over $\mathbb{Q}$. Hence, if $n$ is not of the above form, then any integral point with $X$ dividing $n$ will be of infinite order and hence the rank will be at least $1$.
For all primes $p$ dividing $n$, the curve has split multiplicative reduction with $3\cdot \text{ord}_p(n)$ components.
Maybe a descent via three-isogeny could help.