This is only a partial answer, which I shall delete if I find a better one. Every pair $(k,d)$ of the form
$$d=\frac12\phi(n),\qquad k={\rm card}(\frac{n}{6}\le j \le\frac{n}{2},j\wedge n=1)$$
is OK: take the cyclotomic polynomial $\Phi_n$ and form the irreducible polynomial $P_n\in{\mathbb Z}[X]$ defined by
$$\Phi_n(t)=t^{\frac{n}{2}}P_n\left(t+\frac1t\right).$$
The roots of $P_n$ are the numbers $2\cos\frac{2j\pi}{n}$ with $j\wedge n=1$, smaller than $1$ if and only if $\frac{n}{6}\le j \le\frac{n}{2}$.

If instead $k=d-1$, take any Salem number. **Edit** (after Nikita's comment below): One may take the companion matrix of $X^d-X^{d-1}-\cdots-X-1$. Its only root of modulus greter than $1$ is a Pisot number, also called a multinacci number. If $d=2$, this is just the golden rule, at the basis of the Fibonacci sequence, hence the `word' *multinacci*.