Firstly, notice that $n_k = 13\cdot10^{k - 2} + \frac{10^{k - 2} - 1}{3}$ , also $n_k = \frac{399\dots3}{3} = \frac{4\cdot10^{k - 1} - 1}{3}$. If $k$ odd we have $k = 2x + 1 \implies n_k = \frac{(2^{2x} - 1)(2^{2x} + 1)}{3}$, which is impossible since $n_k$ is prime. If $k$ even, $8 \mid k(k + 2)$, so we need to prove that $k$ can't be $2 \mod{3}$. If $k \equiv 2\mod{3}$, $k \equiv 2\mod{6}$, so say $k = 6x + 2$, replacing this gives $ n_k = 13\cdot10^{6x} + \frac{10^{6x} - 1}{3} $ Notice that $10^6 - 1 \mid 10^{6x} - 1$ since $10^{6x} \equiv 1\mod{10^6 - 1}$. Note that $13 \mid 10^6 - 1$, so this means $13 \mid n_k$, impossible, so $k \neq 2\mod{3}$, hence we are done.