1. Define the problem and notation:
We are given functions $f: \mathbb{N} \to \mathbb{N}$ and $g: \mathbb{N} \to \mathbb{N}$ such that $g$ assumes infinitely many values. We need to find all positive integers $k$ for which the equation
$$f^{g(n)}(n) = f(n) + k$$
holds for every positive integer $n$. Here, $f^i$ denotes the function $f$ applied $i$ times.

2. **Introduce the assertion $P(n)$:**
Let $P(n)$ denote the assertion $f^{g(n)}(n) = f(n) + k$.

3. Sublemma 01:
There exists $m_0$ such that $f(m) = f(n) + ak$ for any $n, a \in \mathbb{N}$.

Proof:
We use induction. Notice that there exists an integer $m_1$ such that $f(m_1) = f(n) + k$. Consider $P(m_1)$, we then have $m_2$ such that $f(m_2) = f(m_1) + k = f(n) + 2k$. Continuing this by induction, we get our desired result.

4. Sublemma 02:
For any integer $\ell, m$, we have $f^{\ell}(f(n) + mk) = f^{\ell + 1}(n) + mk$.

Proof:
We prove this by induction. By the standard trick,
$$f(f(n) + k) = f(f^{g(n)}(n)) = f^{g(n)}(f(n)) = f(f(n)) + k$$
Therefore, we can get that if $f^{\ell} (f(n)+k) = f^{\ell + 1}(n) + k$, then
$$f^{\ell + 1}(f(n) + k) = f^{\ell}(f(f(n)) + k) = f^{\ell + 1}(f(n)) + k = f^{\ell + 2}(n) + k$$
Now consider $m_{a - 1}$ such that $f(m_{a - 1}) = f(n) + (a - 1)k = f(m_{a - 2}) + k$ which exists by Sublemma 01, then
$$f^{\ell} (m_a) = f^{\ell - 1}(f(m_{a - 1}) + k) = f^{\ell}(m_{a - 1}) + k \Rightarrow f^{\ell}(f(n) + ak) = f^{\ell + 1}(n) + ak$$

5. Main Claim:
There doesn't exist any value of $\ell > 0$ such that $f^{\ell} (n) = n$ for any $n \in \mathbb{N}$.

Proof:
Define $\mathcal{A}$ as the set of values of $f^{k}(n)$, where $k \in \mathbb{N}$. To prove this, we'll prove that $f^a(n) + ak \in \mathcal{A}$ for any $a \in \mathbb{N}$. We'll prove this by induction. For $n = 1$, this is obvious. Then, plug $P(f^{g(n)}(n))$ and we have
$$f^{g(f(n) + k) + g(n)}(n) = f(f(n)) + 2k$$
Now, suppose $f^{a}(n) + ak \in \mathcal{A}$, then we have $f^{X_{n,a}}(n) = f^a(n) + ak$ for some $n,a$ and $X_{n,a}$ is a variable depending on $n$ and $a$. Then, replace all $n$ with $f^{g(n)}(n) = f(n) + k$, we get
$$f^{X_{f(n) + k,a} + g(n)}(n) = f^a(f(n) + k) + ak = f^{a + 1}(n) + (a + 1)k$$
Therefore, we are done by induction. Now notice that the sequence $\{f^i(k) + ik\}_{i = 1}^{\infty}$ is unbounded. Since they are all members of $\mathcal{A}$, then $|\mathcal{A}| = +\infty$. However, if there exists $\ell$ such that $f^{\ell}(n) = n$, then $|\mathcal{A}|$ is finite, a contradiction.

6. Final steps:
Now, to finish this problem off. Notice that $P(f(n) + ak)$ gives us
$$f^{g(f(n) + ak)}(f(n) + ak) = f(f(n) + ak) + k$$
By our lemma, $f^{g(f(n) + ak) + 1}(n) + ak = LHS = RHS = f(f(n)) + (a + 1)k$, which therefore gives us
$$f^{g(f(n) + ak) + 1}(n) = f(f(n)) + k$$
for any $a \in \mathbb{N}$. However, this gives us
$$f^{g(f(n) + ak) + 1}(n) = f(f(n)) + k = f(f(n) + k) = f(f^{g(n)}(n)) = f^{g(n) + 1}(n)$$
Therefore, we have
$$f^{|g(f(n) + ak) - g(n)|}(n) = n$$
for all $n \in \mathbb{N}$. Therefore, we have $g(n) = g(f(n) + ak)$ for all $a \in \mathbb{N}$. For $k = 1$, there are no such functions obviously as this contradicts $g$ is unbounded.

7. **Construction for $k \ge 2$:**
Consider $f(k^n) = nk + 1$ and $f(nk + 1) = k^n + 2$. For any $a$, which is not a power of $k$, we can let $f(ak) = ak + 2$ and define $f(x) = x + 1$ for any other integers $x$ not mentioned. One can define $g(n)$ to be the smallest $m$ satisfying $f^m(n) = f(n) + k$. One can check that $g$ is unbounded since $g(nk + 1)$ is the number of integers $m$ such that $k^n + 2 \le m \le k^{n + 1}$ and $k \nmid m - 1$.

The final answer is $\boxed{\text{all } k \ge 2}$