The required numbers are $n_1 = n_2 = \dots = n_k = 1$.

For every integer $m > 1$, let $p(m)$ denote the least prime divisor of $m$. We show that, if $m$ and $\ell$ are integers greater than $1$, and $m \mid 2^{\ell} - 1$, then $p(m) < p(\ell)$. Since $p(m)$ is odd, $p(m) \mid 2^{p(m)-1} - 1$, and since $p(m) \mid 2^{\ell} - 1$, it follows that $p(m) \mid 2^{\gcd(\ell, p(m)-1)} - 1$. Notice that $\gcd(\ell, p(m)-1) > 1$, since $p(m) > 1$, to infer that $\ell$ has a prime divisor not exceeding $p(m) - 1$, and conclude thereby that $p(m) < p(\ell)$.

Suppose now, if possible, that $n_1 > 1$. Then $n_k > 1$, so $n_{k-1} > 1$, and so on and so forth all the way down to $n_2 > 1$. Hence $p(n_1) < p(n_2) < \dots < p(n_k) < p(n_1)$ which is a contradiction. Consequently, $n_1 = 1$, so $n_2 = 1$, and so on and so forth all the way up to $n_k = 1$.