You are asking for rational points in the parabola $$x^2-4y+1=0.$$ It is easy to find one of such points, say $(0, 1/4)$. Take a line of rational slope $q$ that passes through $(0,1/4)$ and compute the intersection between the line and the parabola.
The line is $y=qx+\frac14$. Put this is the parabola equation to get $x^2-4qx=0$. Since you don't want the point where $x=0$ (a line and a parabola generally intersect at two points) the other point must have $x=4q$. This also shows the $y$ of the new point is $4q^2+\frac14$.
That is, take any rational $q$. Then $(x,y)=\left(4q, 4q^2+\frac 14\right)$ is a solution of our desired equation. Note that in the original problem, $m/n$ is our $y$. A more careful analysis shows these are actually all the solutions.