= Solution
Put $A=R/(p^k)$. A <principal ideal domain> is a <unique factorization domain>, so every nonzero element of this <quotient ring> can be written
$$
\overline a=\overline{p}^{\,e}\overline s,
\qquad 0\leq e<k,\qquad p\nmid s.
$$
Indeed, remove the largest power of the <prime element> $p$ from a representative that is not divisible by $p^k$. Since $s$ and $p^k$ have <greatest common divisor> $1$ up to a <unit>, <Bézout's identity> in the <principal ideal domain> gives $xs+yp^k=1$. Thus $\overline s$ is a <unit> in $A$.
Consequently every nonzero <principal ideal> in $A$ is generated by one of $1,\overline p,\ldots,\overline p^{\,k-1}$. Include the zero <ideal> by assigning it exponent $e=k$. These <ideals> are distinct: their inverse images in $R$ are $(1),(p),\ldots,(p^k)$, and equality of two different powers would, by cancellation in the <integral domain>, make the nonunit $p$ a <unit>.
Now suppose $(\overline a)=(\overline b)$. If this is the zero <ideal>, both generators are zero and are <associate elements>. Otherwise their exponents agree, so write
$$
\overline a=\overline p^{\,e}\overline s,\qquad
\overline b=\overline p^{\,e}\overline t
$$
with $\overline s,\overline t$ both <units>. Then
$$
\overline a=(\overline s\,\overline t^{-1})\overline b,
$$
and the multiplier is a <unit>. Thus
$$
\boxed{R/(p^k)\text{ is an associator ring for every }k\geq1.}
$$
The exponent detects the <principal ideal>, while its remaining factor is invertible; this is why <zero divisors> cause no obstruction here.
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