If $(x+1)P(x-1) - (x-1)P(x) = k$, then also $(x+2)P(x) - xP(x+1) = k$. Subtract the two, to get $(2x+1)P(x) - xP(x+1) - (x+1)P(x-1) = 0$, or $(x+1)(P(x) - P(x-1)) = x(P(x+1)-P(x))$. Denoting $\Delta(x) = P(x+1)-P(x)$, that writes $(x+1)\Delta(x-1) = x\Delta(x)$.
Thus $\Delta(-1) = 0$, so $\Delta(x) = (x+1)Q(x)$, and then $Q(x-1) = Q(x)$, forcing $Q(x) = C$ (constant), thus $\Delta(x) = C(x+1)$. But it is known and easy to prove that $\deg \Delta = \deg P - 1$, so $P(x) = ax^2+bx+c$. Then $C(x+1) = \Delta(x) = P(x+1)-P(x) = a(x^2+2x+1) + b(x+1) + c - ax^2-bx-c$, and identifying the coefficients we obtain $a=b =C/2$, thus the general form is $P(x) = sx^2 + sx + t$, for arbitrary coefficients $s,t$.