For $n\geq 2$, consider the polynomial $f_n(x):=x^n-(x-1)^n-1$.
Let $m\neq n$, and let $d=d(m,n):=gcd(f_n,f_m)$. Via extensive computer calculations, I am convinced that the following must hold:
When either $m$ or $n$ (or both) is even $d=x-1$ If $m$ and $n$ are both odd, then $=x(x-1)$ when $3$ divides $m$ or $n$ (or both). If $(m,n)\equiv (1,1)\pmod 6$, then $d=(x^2-x+1)^2x(x-1)$; if $(m,n)\equiv (-1-1)\pmod 6$, then $d=(x^2-x+1)x(x-1)$, and if $(m,n)\equiv (\pm 1, \mp 1)\pmod 6$ then $d=(x^2-x+1)(x-1)$.
I might have switched some of the answers, but the point is that there is periodicity modulo $6$.
So far I could only show that $d$ is a multiple of what I claim to be the correct answer.