The question is:
Find all pairs $(n,m)\in\mathbb{N}^2$ such that $$2(n-1)n(n+1)(n+2)=(m-3)(m+3)$$
I checked all $n<10000$ and only got $n=1$ and $n=4$ with their corresponding $m$, so I suspect those are the only solutions. I tried using quadratic residues, or the fact that the expression is factored, but no luck. I even tried solving for $n$, and using the fact that that has to be an integer, but that led to $$((2n+1)^2-5)^2=8N^2-56$$ which wasn't that useful either. Can anyone help me out here? Thanks!
The nonnegative integer solutions of the Pell equation are $\pmatrix{m\cr a\cr} = M^k \pmatrix{3\cr 1\cr}$ and $M^k \pmatrix{5\cr 3\cr}$ for $k \ge 0$, where $M = \pmatrix{3 & 4\cr 2 & 3\cr}$.
In order for $a = n^2 + n - 1$, we need $5+4a$ to be a square. It appears that the only cases where this is true are $M^0 \pmatrix{3\cr 1\cr} = \pmatrix{ 3\cr 1\cr}$, corresponding to the solution $(m=3,n=1)$, and $M^1 \pmatrix{5\cr 3\cr} = \pmatrix{27\cr 19\cr}$, corresponding to $(m= 27, n= 4)$. There are no other solutions for $k \le 10000$ (corresponding to $n \le 1.03 \times 10^{3828})$.
However, I don't have a proof that these are all the solutions.
I might note that your equation is an elliptic curve, with Weierstrass form $s^3- \dfrac{268}{3} s-\dfrac{7040}{27}+t^2 = 0$ according to Maple. An expert in elliptic curves (not I) might be able to prove that there are only the two positive integer solutions.