= Solution
The <product of ideals> $AB$ consists of finite sums $\sum_ja_jb_j$ with $a_j\in A$, $b_j\in B$. <Ideal> divisibility $P\mid A$ means $A=PC$ for an <integral ideal> $C$; in this ring it is equivalent to $A\subseteq P$. The <ideal norm> is $N(A)=|\mathcal O_K/A|$, equivalently its additive lattice index.
If the <prime ideal> $P$ divides $AB$ but does not divide $A$, choose $a\in A\setminus P$. For every $b\in B$, $ab\in AB\subseteq P$, so primality gives $b\in P$. Thus $B\subseteq P$, proving $P\mid B$.
Every nonzero proper <ideal> $A$ is contained in a maximal <ideal> $P$, which is prime. Then $P\mid A$, so $A=PB$. <Norm> multiplicativity gives $N(B)=N(A)/N(P)<N(A)$ because $N(P)\ge2$. Induction on the positive <integer> $N(A)$ gives a prime-<ideal> factorization, ending at <norm> one. For uniqueness, a prime in one factorization divides some prime in the other. Every nonzero <prime ideal> is maximal, since its quotient is a finite integral domain and hence a <field>. The two primes therefore coincide. Cancel that common factor using the permitted cancellation law, and continue. This proves uniqueness up to order.
The finite additive <group> $\mathcal O_K/A$ has order $N(A)$, so that <integer> annihilates the class of $1$. Therefore
$$
\boxed{N(A)\in A.}
$$
In particular an <ideal> of <norm> $m$ contains $(m)$. Such ideals correspond to certain additive subgroups of the finite <group> $\mathcal O_K/(m)$, so there are finitely many. Finiteness of these quotients follows from an <integral basis>; a nonzero <ideal> contains a nonzero <principal ideal>, whose index is the absolute <field norm> of its generator.
For the computation put $s=\sqrt{-6}$. The <Kummer-Dedekind theorem> says that when $p$ does not divide the power-<basis> index, a factorization $\bar f=\prod_i\bar g_i^{e_i}$ gives $(p)=\prod_i(p,g_i(s))^{e_i}$, with each factor having <norm> $p^{\deg g_i}$. The given full power <basis> has index one. Since $X^2+6\equiv X^2\pmod2$ and $X^2+6\equiv(X-1)(X+1)\pmod7$,
$$
\boxed{(2)=P^2,\quad P=(2,s);\qquad
(7)=QQ',\quad Q=(7,s-1),\ Q'=(7,s+1).}
$$
Here $N(P)=2$ and $N(Q)=N(Q')=7$. A <prime ideal> dividing an <ideal> of <norm> $98=2\cdot7^2$ lies above $2$ or $7$, and these are all the prime ideals above them. Unique factorization and <norm> multiplicativity therefore leave exactly
$$
\boxed{PQ^2,\qquad PQQ'=(7)P,\qquad P(Q')^2.}
$$
They are distinct by their prime-<ideal> valuations, and each has <norm> $98$.
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