= Solution
First let $I=\mathfrak p^e$ for a prime ideal $\mathfrak p$ and $e\geq1$. If $\alpha\beta\in\mathfrak p^e$ but $\alpha\notin\mathfrak p^e$, then in the prime-ideal factorization valuations
$$
v_{\mathfrak p}(\alpha)+v_{\mathfrak p}(\beta)\geq e,
\qquad
v_{\mathfrak p}(\alpha)<e.
$$
Thus $v_{\mathfrak p}(\beta)>0$, so $\beta\in\mathfrak p$ and $\beta^e\in\mathfrak p^e=I$. Hence every prime-ideal power is primary.
Conversely, the radical of any <primary ideal> is prime. To see this, suppose $ab\in\sqrt I$ and $a\notin\sqrt I$. For some $m$, $a^mb^m\in I$, while $a^m\notin I$. Primaryness gives $(b^m)^k\in I$ for some $k$, and hence $b\in\sqrt I$.
For nonzero $I$, unique factorization of ideals gives
$$
I=\mathfrak p_1^{e_1}\cdots\mathfrak p_r^{e_r},
$$
so
$$
\sqrt I=\mathfrak p_1\cdots\mathfrak p_r.
$$
This radical is prime only when $r=1$. Therefore a nonzero primary ideal is $\mathfrak p^e$. The zero ideal is itself prime because $\mathcal O_K$ is a domain, so it is its own first power. Consequently
$$
\boxed{I\text{ is primary}\quad\Longleftrightarrow\quad
I\text{ is a power of a prime <ideal>}.}
$$
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