Inspired by Satyajit Mohanty, but this is one is a lot easier!

From the equation $(x+1)P(x-1)=(x-1)P(x)\:\:\:(*),$ we obtain that 0, and -1 are roots of $P(x)$ . Therefore, $P(x)=x(x+1)Q(x)$ where $Q(x)$ is a polynomial. By substituting this expression for $P(x)$ into $( * )$ we obtain the equation $(x+1)(x-1)x Q(x-1)=(x-1)x(x+1)Q(x).$ Dividing both sides by $(x+1)(x-1)x$ we get that $Q(x-1)=Q(x)$ . The latter implies that $Q(x)=Q(x+n)$ for all real values of x and any integer $n$ . Then $Q(x)$ must be a constant polynomial. Indeed, if $Q(x)$ were not constant then for an arbitrary $x$ , $Q(x)= \lim_{n\rightarrow \infty}Q(x+n)=\pm \infty,$ that would be a contradiction. Assuming that $Q(x)$ is the constant $c$ we obtain that $P(x)= c x(x+1)$ and therefore the maximum possible degree of $P(x)$ would be 2.

By substituting $x=0$ and $x= 1$ to the equation, we obtain $P(0) = P(-1) = 0$ , so P(x) has degree at least 2. Thus, we have $P(x) = x(x+1)A(x)$ for some polynomial $A(x)$ . Substituting this to the equation, we get $(x+1)(x-1)(x)A(x-1) = (x-1)(x)(x+1)A(x)$ . This implies that $A(x-1) = A(x)$ for all real numbers $x$ . THis means that $A(x)$ is constant, so the maximum degree of $P$ is 2.

Rewrite as P(x)/P(x-1)=(x+1)/(x-1). There will be common roots cancelled and by inspection, the maximum degree of P(x) must be 2