To determine for which $(n, m)$, the only $(n, m)$-good set is $\mathbb{N}$, we need to examine the conditions given in the problem and their consequences.

A set $S$ of positive integers is called $(n, m)$-good if:

1. $m \in S$,
2. For all $a \in S$, all divisors of $a$ are also in $S$,
3. For all distinct $a, b \in S$, $a^n + b^n \in S$.

We are tasked with finding values of $n$ such that the only set fulfilling these conditions is $\mathbb{N}$, the set of all positive integers.

### Step-by-Step Analysis:

1. Condition 1 ensures that the element $m$ is included in the set $S$.

2. Condition 2 implies a closure property under the division: all divisors of any element in $S$ must also be in $S$.

3. Condition 3 needs more attention, as it extends the set whenever two distinct elements $a, b$ are present. If the set $S$ is not already $\mathbb{N}$, adding element $a^n + b^n$ should eventually force $S$ to include all positive integers.

- For $n$ odd:
- Consider any positive integer $k$. Choose $a = k$ and $b = 1$. Then $a^n + b^n = k^n + 1 \equiv k + 1 \pmod{n}$. Due to the successive increments with odd powers, all larger numbers are eventually included in $S$. Iterating this process leads $S$ to include all integers, thereby proving $S = \mathbb{N}$.

- For $n$ even:
- The element $a^n + b^n$ with $n$ even can have gaps in the integers it produces from elements of $S$. Specifically, $a = 1$ and $b = 1$ result in $2 \cdot 1^n = 2$, which doesn't necessarily generate all integers, maintaining the possibility of a restricted set.

### Conclusion:

The critical factor is whether $n$ is odd or even. When $n$ is odd, the condition $(3)$ progressively generates all earlier and further numbers from any starting integer $m$, thus $(n, m)$-good set becomes $\mathbb{N}$. Conversely, when $n$ is even, this cascading effect does not occur universally.

Therefore, the given set $S$ is $\mathbb{N}$ if and only if $n$ is odd.

$$\boxed{\text{The only } (n, m)\text{-good set is } \mathbb{N} \text{ if and only if } n \text{ is odd.}}$$