see, by checking divisibility of squares by $ 3$ we obtain remainder as $ 1$ or none. checking the above equation, either both $ 4m$ and $ n$ need to be multiples of $ 3$ or then w.l.g. $ m\equiv 1mod3$ and $ n\equiv 2mod3$. in the first case the power 4 in LHS is high, compulsing $ m$ and $ n$ to be multiples of $ 9$ and so on, ultimately solving only at values $ >N$. the second case is possible...and needs to be worked.. ll be back soon

My solution: Denote $ (m + n)^2 = x$ and $ mn = y$. Then the original equation is equivalent to $ x^2 - x = y^2 + 4y$. Solving this equation wrt to $ y$, one gets $ y = - 2 \pm \sqrt {x^2 - x + 4}$. Hence $ x^2 - x + 4$ has to be a square. But $ (x - 1)^2 < x^2 - x + 4 < x^2$ for $ x\ge 5$. Hence there are just 5 easy cases left.

What are the solution s