Example 16 Find all real-coefficient binary polynomials $f(x, y)$ of degree $n$, such that for any real numbers $x, y$, we have $f(x+1, y+1)=f(x, y)$.
To solve the problem of reducing a binary polynomial to a unary polynomial, let
$$
x=u+v, y=u-v .
$$

Then $u=\frac{1}{2}(x+y), v=\frac{1}{2}(x-y)$.
Let $f(x, y)=f(u+v, u-v)=F(u, v)$.
$F(u, v)$ is also an $n$-degree binary polynomial with real coefficients. Using (1) and (3), we get
$$
f(x+1, y+1)=f[(u+1)+v,(u+1)-v]=F(u+1, v) \text {. }
$$

Using the given conditions, we have $F(u+1, v)=F(u, v)$.
From the above equation, for any positive integer $n \geqslant 2$, we have
$$
F(u+n, v)=F[u+(n-1), v]=F[u+(u-2), v]=\cdots=F(u, v) \text {. }
$$

Since $x, y$ are arbitrary real numbers, from (2), $u, v$ can take any real values.
For any two real numbers $a, b$, let $g(u)=F(u, b)-F(a, b)$,
where $a, b$ are fixed, and $g(u)$ is a unary polynomial in $u$ with real coefficients, at most $n$ times. Using (5) and (6), for any positive integer $n$, we have
$$
g(a+n)=F(a+n, b)-F(a, b)=0 .
$$

Therefore, the polynomial $g(u)$ has infinitely many real roots $a+1, a+2, a+3, \cdots, a+n, \cdots$, thus the polynomial $g(u)$ is identically zero. Using (7), for any real number $u$, we have
$$
F(u, b)=F(a, b) .
$$

Taking $a=0, b=v$, from (9), we have $F(u, v)=F(0, v)$.
Since $u, v$ are any two real numbers, using (2), (3), and (11), for any real numbers $x, y$, we have
$$
f(x, y)=F(u, v)=F(0, v)=F\left[0, \frac{1}{2}(x-y)\right] .
$$

Clearly, $F\left[0, \frac{1}{2}(x-y)\right]$ is a polynomial in $x-y$ with real coefficients. Since $f(x, y)$ is an $n$-degree binary polynomial with real coefficients, we have
$$
f(x, y)=F\left[0, \frac{1}{2}(x-y)\right]=\sum_{k=0}^{n} a_{k}(x-y)^{k},
$$

where $a_{n}$ is any non-zero real number, and $a_{k}(0 \leqslant k \leqslant n-1)$ are any real numbers.