Example 2 Let $a \in \mathbf{N}_{+}$. Find the function $f: \mathbf{N}_{+} \rightarrow \mathbf{R}$, such that for any $x, y \in \mathbf{N}_{+}, x y>a$, we have
$$
f(x+y)=f(x y-a) .
$$
Solution: For any $t \in \mathbf{N}_{+}$, since $t=(t+a)-a$ and $1+(t+a)=a+1+t$, then
$$
\begin{array}{l}
f(t)=f((t+a) \times 1-a) \\
=f(1+t+a)=f((a+1)+t) \\
=f((a+1) t-a) \\
=f((a+1)(t-1)+1) \\
=f((a+1)(t-1)-a)=\cdots \\
=f((a+1) \times 1-a)=f(1) .
\end{array}
$$

Therefore, the function sought is a constant function, and $f(x)=f(1)$ (a constant).