With the transformation $X = -n/x$ and $Y= ny/x$, the curve becomes isomorphic to the Weierstrass model 
$$ E_n\colon \  \ Y^2 - X\ Y - n\ Y = X^3.$$ 
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 $1$. 

(Edit:)
Now, I have a reason to believe that the number **is** bounded. As pointed out by Felipe Voloch in
<a href="https://mathoverflow.net/questions/50661/unboundedness-of-number-of-integral-points-on-elliptic-curves">this question</a>, the <a href="http://www.springerlink.com/content/xgbvqxm4383nwhqu">paper</a> by Abramovich shows that:

*if the conjecture by Lang and Vojta about rational poitns on varieties of general type holds, then the number of solutions is bounded as $n$ varies.* 

One has just to note that the equation $E_n$ is in fact minimal and that the curve $E_n$ is semistable for all $n$. For all primes $p$ dividing $n$, the curve has split multiplicative reduction with $3\cdot \text{ord}_p(n)$ components. For all primes $p$ dividing $27n+1$, the reduction can be shown to be multiplicative, as well.

Maybe a descent via three-isogeny could help to give an upper bound on the rank.