Let's start find those $n$ for which $m$ is an integer.

Firstly, the pair $(\frac{m}{2}-1, n)$ must be a solution of Pell's equation $x^2-28y^2=1$ whose fundamental solution is $(x_1, y_1)=(127, 24)$.

Hence $\frac{m}{2}-1+n\sqrt{28}=(127+24\sqrt{28})^k$ for some $k\in\mathbb{N}$.

Now we have $m=2+(127+24\sqrt{28})^k+(127-24\sqrt{28})^k=A^2$

where $A=(8+3\sqrt{7})^k+(8-3\sqrt{7})^k$ is an integer.