Let $\mathcal P_{{\rm fin},0}(\mathbf N)$ be the monoid of non-empty, finite subsets of $\mathbf N$ with $0 \in \mathbf N$ equipped with the operation $$\mathcal P_{{\rm fin},0}(\mathbf N) \times \mathcal P_{{\rm fin},0}(\mathbf N) \to \mathcal P_{{\rm fin},0}(\mathbf N): (X, Y) \mapsto \{x+y: x \in X,\, y \in Y\},$$ which we'll denote, as usual, with the symbol "$+$".

We refer to a set $A \in \mathcal P_{{\rm fin},0}(\mathbf N)$ as an atom if (i) $A \ne \{0\}$ and (ii) $A \ne X+Y$ for all $X, Y \in \mathcal P_{{\rm fin},0}(\mathbf N) \setminus \bigl\{\{0\}\bigr\}$: Note that $\{0\}$ is the identity (and, in fact, the only unit) of $\mathcal P_{{\rm fin},0}(\mathbf N)$, so this is nothing but a special instance of the general notion of atom (or irreducible element) for an arbitrary monoid.

With this in mind, fix $X \in \mathcal P_{{\rm fin},0}(\mathbf N)$. We take ${\sf L}(X) := \{0\}$ if $X = \{0\}$; otherwise, ${\sf L}(X)$ is the set of all $k \in \mathbf N^+$ for which there exist atoms $A_1, \ldots, A_k \in \mathcal P_{{\rm fin},0}(\mathbf N)$ such that $X = A_1 + \cdots + A_k$: In factorization theory, ${\sf L}(X)$ would be called the set of lengths of $X$ (relative to the atoms of $\mathcal P_{{\rm fin},0}(\mathbf N)$), which explains the title.

Q. Is it true that $|{\sf L}(nX)| \to \infty$ as $n \to \infty$ unless $X = \{0\}$?

Of course, it's enough to show that the answer is yes in the special case when $X$ is an atom, though I'm not so sure that this makes things any easier. What is perhaps more interesting is that a positive answer is equivalent to showing that there exists $n \in \mathbf N^+$ such that $|\mathsf L(nX)| \ge 2$: This comes as a consequence of a general, simple fact about sets of lengths in factorization theory.

Incidentally, it's probably worth mentioning that $\mathcal P_{{\rm fin},0}(\mathbf N)$ is a (reduced) BF-monoid, that is, $1 \le |\mathsf L(X)| < \infty$ for every non-zero $X \in \mathcal P_{{\rm fin},0}(\mathbf N)$.