Your reasoning is correct but you need to formalize your argument.
Here is a more rigorous but not so epsilon-delta-ish way to do it:
The set is clearly bounded above by $1$ and:
$$\lim_{n \to \infty} \frac{n}{n+m} = 1$$
So the supremum is $1$. If it's attained then $n+m = n$ for some $n,m \neq 0$, which is impossible. So the maximum doesn't exist.
The set is bounded below by $0$ and:
$$\lim_{m\to \infty}\frac{n}{n+m} = 0$$
So the infimum is $0$ and it's clearly not attained. So the minimum also doesn't exist.
To do it epsilon-delta style, let $\epsilon >0$. We have to prove that we can find $n,m>0$ such that:
$$\frac{n}{n+m} > 1 - \epsilon$$
i.e.
$$\frac{m}{n+m} < \epsilon$$
Take $m=1$. By the Archimedean property there exists $n$ such that $\frac1{n+1}<\epsilon$. So we are done.
Once again for $\epsilon >0$, we have to find $n,m$ such that:
$$\frac{n}{n+m} < \epsilon$$
Repeat the above argument, swapping the roles of $n$ and $m$.