Two different sums!

$\begin{aligned} & 1+2+\cdots + M = 1^2+2^2+\cdots + N^2 \\& \frac{M(M+1)}{2} = \frac{N(N+1)(2N+1)}{6} \\& 3M^2 +3M -N(N+1)(2N+1) = 0 \\ & M = \frac{-3 + \sqrt{3 (8N^3+12N^2 +4N+3)}}{6} \end{aligned}$ Now noting the discriminant of above equation $\begin{aligned} & D = 8N^3+ 12N^2 +4N +3 \\& = 8\left(N+\frac{1}{2}\right)^3- 2\left(N+\frac{1}{2}\right)+3 \\ & = X^3 -X +3 \phantom{cccccc} \text{let x} = 2\left(N+\frac{1}{2}\right)\end{aligned}$ It is vivid that discriminant $D$ in the above equation will be perfect square iff $\begin{aligned} & D = 3Y^2 \\ & X^3 -X+3 = 3Y^2 \cdots(a)\end{aligned}$ The solution of equations $a$ has been provided by Saburô Uchiyama and one can Google it as $\text{ Solution of a Diophantine equation by saburô Uchiyama}$

So the solution for $M,N$ can be interred as $(10,5)$ , $(13,6)$ , $(645,85)$

The actual answer to the original(unedited) question was 2 , where both m=n=1 . But assuming both m and n greater than 3 , we will get (10,5)