It is difficult to impose conditions only on the spectrum of a bounded linear operator that imply the operator in question is compact. There are none that I am aware of and the following example shows that it is not reasonable to expect that there are such conditions.
Define $\tau \colon L^{2}(\mathbb{R}) \to L^{2}(\mathbb{R})$ by $(\tau f)(x) := f(x-1)$. Then $\tau$ is a well-defined isometric isomorphism on $L^{2}(\mathbb{R})$. Define $\phi \colon \mathbb{R} \to \mathbb{R}$ by $\phi (x) := \exp (-x^{2})$. As $\phi \in L^{\infty}(\mathbb{R})$ and $\phi$ is not equal to zero almost everywhere, we have that $M_{\phi}\colon L^{2}(\mathbb{R}) \to L^{2}(\mathbb{R})$ defined by $M_{\phi}f := \phi f$ is a bounded linear operator on $L^{2}(\mathbb{R})$ that is not compact. Now define $A := M_{\phi} \circ \tau$. Then $A$ is a bounded linear operator on $L^{2}(\mathbb{R})$ that is not compact. Some simple calculations show that
\begin{equation}
\|A^{n}\| = \sup_{x\in\mathbb{R}} \Big|\exp \Big(-\sum_{k=0}^{n-1} (x - k)^{2} \Big)\Big| \leq \exp (- \tfrac{(n-1)^{2}}{4})
\end{equation}
for each $n\in\mathbb{N}$ and consequently that $\lim_{n\to\infty} \|A^{n}\|^{\tfrac{1}{n}} = 0$. Hence we have $\sigma (A) = \{0\}$.
This provides an example of a bounded linear operator on a separable Hilbert space that is not compact and yet has spectrum equal to $\{0\}$. In particular, the spectrum of this non-compact operator satisfies the three properties from your question.
However, if, in addition to the assumptions on the spectrum, the operator is a normal (or self-adjoint if the Hilbert space is real) operator on a Hilbert space, then it follows that the operator is compact. See here for a proof.
