Find the number of pairs(m,n) of integers which satisfy the equation

Similar to @Brian Moehring solution, we may take both sides, mod 3

$m^3-m \equiv 1 (mod \ 3)$

Also $GCD(m,3)=1$ , for if $m$ is a multiple of $3$ , then the RHS of the main equation should be divisible by $3$ , which is not possible.

Since $GCD(m,3)=1$ , using Fermat little theorem,

$m^3-m \equiv 1 (mod \ 3) \implies m-m\equiv 0 \equiv 1 (mod \ 3)$

which is a contradiction to the existence of any solution to the equation.

The given equation is equivalent to $3\left(2\binom{m+1}{3} + 2m^2 + 2m - 9n^3 - 3n^2 - 3n\right) = 1$

However, if $m,n$ are integers, then the left side is an integral multiple of $3$ while the right side isn't. Therefore there are $\boxed{0}$ integral solutions $(m,n)$