Let $P(x)=a_nx^n+\cdots+a_0x^0$ with $a_n\ne 0$. Comparing the coefficients of $x^{n+1}$ on both sides gives $a_n(n-2m)(n-1)=0$, so $n=1$ or $n=2m$.
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 $P(x)=tx$ for $t\in \mathbb{R} $.
Now, assume that $n=2m$. The polynomial $xP(x+1)-(x+1)P(x)=(n-1)a_nx^n+\cdots$ has degree $n$, and therefore it has at least one (possibly complex) root $r$. If $r\ne \{0,\,-1\}$, define $k=\dfrac{P(r)}{r}=\dfrac{P(r+1)}{r+1}$. If $r=0$, let $k=P(1)$. If $r=-1$, let $k=-P(-1)$. We now consider the polynomial $S(x)=P(x)-kx$. It also satisfies (1) because $P(x)$ and $kx$ satisfy it. Additionally, it has the useful property that $r$ and $r+1$ are roots.
Let $A(x)=x^3-mx^2+1$ and $B(x)=x^3+mx^2+1$. 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 $a\ge 0$ and $b\ge 1$ be such that $r-a,\,r-a+1,\,\cdots, r,\,r+1,\.\cdots,\,r+b-1,\,r+b$ 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 divisior $C(x)$ 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 $n=2m$.
Therefore $B(x)=A(x+a+b)$. Writing $c=a_b\ge 1$, we compute
$0=A(x+c)-B(x)=(3c-2m)x^2+c(3c-2m)x^2+c(3c-2m)x+c^2(c-m)$
Then we must have $3c-2m=c-m=0$, which gives $m=0$, a contradiction. We could conclude that $f(x)=tx$ is the only solution.