For $n=2k$ the equation writes $m-1 = (m^2-k)(m^2+k+1)$, and since $m-1 \leq m^2+k+1$ that leads to the unique solution $(n,m) = (2,1)$.

$n(n+2)=4m^4+4m^2-4m+4$ $(2m^2)^2<(n+1)^2=4m^4+4m^2-4m+5<(2m^2+1)^2$ for every $m>1$ For $m=1$ take the solution $(m,n)=(1,2)$

Good idea!