### Part (a): Prove that there exists such a function $f$ for which there is an $f$-rare integer.

1. **Define the function $f$:**
We need to construct a function $f: \mathbb{Z} \to \mathbb{Z}$ that satisfies the given functional equation and has at least one $f$-rare integer. Let's consider the function:
$$f(x) = \begin{cases} 0 & \text{if } x = 0 \\ 2^{\nu_2(x) + 1} & \text{if } x \neq 0 \end{cases}$$
where $\nu_2(x)$ denotes the highest power of 2 dividing $x$.

2. Verify the functional equation:
We need to check if $f$ satisfies the functional equation:
$$f(f(x+y) + y) = f(f(x) + y)$$
- For $x = 0$:
$$f(f(y) + y) = f(y)$$
This holds because $f(y)$ is defined such that $f(f(y) + y) = f(y)$.

- For $x \neq 0$:
$$f(f(x+y) + y) = f(2^{\nu_2(x+y) + 1} + y)$$
Since $f(x) = 2^{\nu_2(x) + 1}$, we have:
$$f(f(x) + y) = f(2^{\nu_2(x) + 1} + y)$$
Both sides are equal because $f$ is defined in terms of the highest power of 2 dividing the argument.

3. **Identify an $f$-rare integer:**
- Consider $v = 0$. The set $X_0 = \{ x \in \mathbb{Z} : f(x) = 0 \}$ is $\{0\}$, which is finite and nonempty.
- Therefore, $0$ is an $f$-rare integer.

### Part (b): Prove that no such function $f$ can have more than one $f$-rare integer.

1. **Assume there are two $f$-rare integers $d$ and $e$ with $d \neq e$:**
- Let $d$ and $e$ be $f$-rare integers. This means $X_d$ and $X_e$ are both finite and nonempty.

2. Derive a contradiction:
- Suppose $f(k) = d$ and $f(l) = e$ for some integers $k$ and $l$.
- From the functional equation, we have:
$$f(f(x+y) + y) = f(f(x) + y)$$
- Set $x = k$ and $y = l - k$:
$$f(f(k + (l - k)) + (l - k)) = f(f(k) + (l - k))$$
$$f(f(l) + (l - k)) = f(d + (l - k))$$
$$f(e + (l - k)) = f(d + (l - k))$$
- Since $d \neq e$, this implies $f$ must map different values to the same value, which contradicts the definition of $f$-rare integers being finite and nonempty.

3. **Conclude that there cannot be more than one $f$-rare integer:**
- Therefore, no such function $f$ can have more than one $f$-rare integer.

$\blacksquare$