Let $P(x)=a_{n} x^{n}+\cdots+a_{0} x^{0}$ with $a_{n} \neq 0$. Comparing the coefficients of $x^{n+1}$ on both sides gives $a_{n}(n-2 m)(n-1)=0$, so $n=1$ or $n=2 m$. If $n=1$, one easily verifies that $P(x)=x$ is a solution, while $P(x)=1$ is not. Since the given condition is linear in $P$, this means that the linear solutions are precisely $P(x)=t x$ for $t \in \mathbb{R}$. Now assume that $n=2 m$. The polynomial $x P(x+1)-(x+1) P(x)=(n-1) a_{n} x^{n}+\cdots$ has degree $n$, and therefore it has at least one (possibly complex) root $r$. If $r \notin\{0,-1\}$, define $k=P(r) / r=P(r+1) /(r+1)$. If $r=0$, let $k=P(1)$. If $r=-1$, let $k=-P(-1)$. We now consider the polynomial $S(x)=P(x)-k x$. It also satisfies (1) because $P(x)$ and $k x$ satisfy it. Additionally, it has the useful property that $r$ and $r+1$ are roots. Let $A(x)=x^{3}-m x^{2}+1$ and $B(x)=x^{3}+m x^{2}+1$. Plugging in $x=s$ into (1) implies that: If $s-1$ and $s$ are roots of $S$ and $s$ is not a root of $A$, then $s+1$ is a root of $S$. If $s$ and $s+1$ are roots of $S$ and $s$ is not a root of $B$, then $s-1$ is a root of $S$. Let $a \geqslant 0$ and $b \geqslant 1$ be such that $r-a, r-a+1, \ldots, r, r+1, \ldots, r+b-1, r+b$ are roots of $S$, while $r-a-1$ and $r+b+1$ are not. The two statements above imply that $r-a$ is a root of $B$ and $r+b$ is a root of $A$. Since $r-a$ is a root of $B(x)$ and of $A(x+a+b)$, it is also a root of their greatest common divisor $C(x)$ as integer polynomials. If $C(x)$ was a non-trivial divisor of $B(x)$, then $B$ would have a rational root $\alpha$. Since the first and last coefficients of $B$ are $1, \alpha$ can only be 1 or -1 ; but $B(-1)=m>0$ and $B(1)=m+2>0$ since $n=2 m$. Therefore $B(x)=A(x+a+b)$. Writing $c=a+b \geqslant 1$ we compute $$0=A(x+c)-B(x)=(3 c-2 m) x^{2}+c(3 c-2 m) x+c^{2}(c-m) .$$ Then we must have $3 c-2 m=c-m=0$, which gives $m=0$, a contradiction. We conclude that $f(x)=t x$ is the only solution.