Find all integer solutions of: $\;\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$ I found the following problem from the 10th Iranian Mathematical Olympiad in Crux Magazine.

Find all integer solutions of $$\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$$

Initially it looked like a typical quadratic problem, however I hit a dead end each time I solve it. 
My methodology is as follows, 
$$\frac{n^2+mn-1}{mn^2}=\frac{3}{4}$$
$$\implies 4n^2+4mn-4 = 3mn^2$$
$$\implies (4-3m)n^2+4m \cdot n-4=0$$
I used the quadratic formula, and got, 
$$n = \frac{-4m \pm \sqrt{(4m)^2-4\cdot(4-3m)\cdot(-4)}}{2(4-3m)}$$
I do the usual algebraic manipulations and drop at, 
$$n = \frac{-2m \pm 2\sqrt{m^2-3m+4}}{4-3m}$$
I am unsure how I go ahead from this. 
Some help would be much appreciated. 
Cheers!

As an alternative to Dr. Sonnhard Graubner's answer on how to find the complete set of solutions from 
$m=\frac{4(n^{2}−1)}{3n^{2}−4n}$
consider the following:
We can factor that expression further to 
$m=\frac{4(n+1)(n-1)}{n(3n−4)}$
That tells us a lot about $n$.  First off, if $n$ is 1, then $m$ must be 0, but that is not a valid solution because the original expression involves a division by $m$.  So $n$ cannot be 1. For the same reason, $n$ cannot be -1.
Since $n$ is an integer, both the numerator and denominator of that expression are integers.  In order for their quotient $m$ to also be an integer, the numerator must be divisible by the denominator.  Since $n\neq\pm1$, both $(n+1)$ and $(n-1)$ must be relatively prime to $n$, so neither $n$ nor any of its prime factors can divide either of these terms.  Therefore, if $m$ is an integer, 4 must be divisible by $n$.  The only candidates are $\pm1$, $\pm2$, and $\pm4$, and we already excluded $\pm1$.  
Substituting $n=2$ gives us $m=3$, so $(m, n) = (3, 2)$ is a valid solution.
Substituting $n=4$ gives us $m=\frac{15}{8}$, which is not an integer and therefore not a valid solution.  Likewise, $n=-2$ yields $m=\frac{3}{5}$ and $n=-4$ yields $m=\frac{15}{64}$.
