= Solution
By the <correspondence theorem for ideals>, an <ideal> of $R/P^n$ has the form $J/P^n$ with $P^n\subseteq J\subseteq R$. If $J\ne R$, apply <unique factorization of ideals in a number field> to $J$. Every <prime ideal> $Q$ dividing $J$ contains $J$ and hence $P^n$. Primality gives $P\subseteq Q$, and maximality gives $Q=P$. Thus $J=P^i$ for some $i\geq1$.
The inclusion $P^n\subseteq P^i$ forces $i\leq n$: if $i>n$, cancellation using the <invertible fractional ideal> $P^n$ would give $R\subseteq P^{i-n}$, which is impossible. The zero <ideal> corresponds to $i=n$, while the unit <ideal> corresponds to $i=0$. Consequently the nonzero proper <ideals> are exactly
$$
\boxed{P^i/P^n,\qquad 1\leq i<n}.
$$
When $n=1$, there are none, in agreement with $R/P$ being a <field>.
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