Assume we have a set of eigenvalues $$0 < \lambda_1 \le \lambda_2 \le \ldots \lambda_n \to \infty,$$ where the only rules I impose are that they are positive and have no limit point.
Does there exists a drum $$D \subset \mathbb{R}^2$$ such that the Laplacian has this set of eigenvalues as its spectrum?
As pointed out by whpowell96, Weyl's law states that
$$ A = 4 \pi \lim_{R \to \infty} \dfrac{N(R)}{R} $$
where $A$ is the area of the drum and $N(R)$ is the number of eigenvalues less than $R$.
We can find counterexamples by looking at sequences of eigenvalues that would imply $A = 0$ or $A = \infty$.
For example, the sequence of eigenvalues $\lambda_n = 2^n$ would have no drum since $N(R) \le \log_2 (R)$ and
$$ A = 4 \pi \lim_{R \to \infty} \dfrac{N(R)}{R} \le 4 \pi \lim_{R \to \infty} \dfrac{ \log_2(R)}{R} = 0$$