This is **not** an answer --- but perhaps someone can fill the gap?

Clearly, we have $d>1$. Denote
$$
n_r=\sum_{i=1}^\infty a_i^r\in\mathbb N.
$$
Choose some $k$ such that
$$
\sum_{i>k}a_i<\frac1{2d}
$$
(all $k$ greater than some $k_0$ fit). Set
$$
P(x)=(x-a_1)\dots(x-a_k)=x^k+p_{k-1}x^{k-1}+\dots+p_0.
$$

Now, look at the rows $m_0,\dots,m_{3k}$ of the matrix $M_{k,s}=[n_{s+i+j}]_{0\leq i,j\leq 3k}$ (for some large $s$). We see that for every $i\geq k$ the elements of
$$
m_i'=m_i+p_{k-1}m_{i-1}+\dots+p_0m_{i-k}
$$
are expressed via the sums of large powers of $a_i$, $i>k$, with bounded coefficients. Thus all elements of $m_i'$ are less than $C_k/(2d)^s$ with some constant $C_k$ depending only on $k$. On the other hand, the entries of $m_0,\dots,m_{k-1}$ do not exceed $\alpha_k kd^{s+k}$ for some constant $\alpha_k$ also depending on $k$ only.

Replacing the $m_i$ with $m_i'$ does not chande $\det M_{s,k}$, so
$$
\det M_{s,k}\leq (3k)!\cdot (\alpha_kkd^{s+k})^k\cdot \left(\frac{C_k}{(2d)^s}\right)^{2k}
$$
which tends to $0$ as $s\to\infty$. On the other hand, $\det M_{s,k}$ is integer, so $\det M_{s,k}=0$ for all sufficiently large $s\geq s_k$.

Now I **would like to** say that this yields $(n_r)$ be a linear recurrent sequence from some moment. Ptifully, this is not true in general, but perhaps some argument may fill this?

Notice that if $(n_r)$ were linear recurrent, the rest is easy. Indeed, we may prove inductively (assuming $a_1\geq a_2\geq \dots$) that all nonzero $a_i$ are the roots of its characteristic polynomial, so they are finitely many.