I claim either $P\equiv 0$, or $m=2023k$ and \[ P(x) = c\prod_{0\le i\le k}(x-2023i), \]where $c\ne 0$ is some constant. First, let $2023\nmid m$. Then, taking $x=0$, we find $P(2023)=0$. Taking $x=2023$, we find $P(2\cdot 2023)=0$. Continuing, we find $P(k\cdot 2023)=0$ for all $k$. So, $P\equiv 0$. Assume now that $2023\mid m$, let $m=2023k$. Taking $x=m$, we find $P(m)=0$. Taking $x=0$, we find $P(2023)=0$. Continuing, we get $P(x)=0$ for $x=2023\cdot i$, $1\le i\le k$. Set $P(x) = \textstyle Q(x)\prod_{0\le i\le k}(x-2023i)$ to find $Q(x+2023)=Q(x)$ for all $x$, so $Q$ is constant.

Sniped lol First assume m is not a multiple of 2023. Then plugging in x=0 we get P(2023)=0, then plugging in x=2023 we get P(4046)=0, and this continues forever, meaning P has infinitely many zeroes so P(x)=0. On the other hand, if m=2023k, this process is only finite, giving us that 2023, 4046, 6069, ..., 2023k are zeroes of P. From here if we let P(x)=(x-2023)(x-4046)...(x-2023k)Q(x) for some polynomial Q, plugging this back in to the original equation almost everything cancels leaving us with Q(x+2023)=Q(x) implying Q is constant, and we are done.