= Solution
Since the <radical of an ideal> $\sqrt{\mathfrak q}$ is $(r)$, some positive power of $r$ belongs to $\mathfrak q$. Choose the least such exponent $k$. Then $(r^k)\subseteq\mathfrak q$ and $r^j\notin\mathfrak q$ for $0\leq j<k$, with $r^0=1$ excluded because the <primary ideal> $\mathfrak q$ is proper.
Take $x\in\mathfrak q$. If $x\notin(r^k)$, choose the largest $j\in\{0,\ldots,k-1\}$ for which $x\in(r^j)$, and write $x=r^ju$. Maximality of $j$ implies $u\notin(r)=\sqrt{\mathfrak q}$: otherwise $x\in(r^{j+1})$, a contradiction, including when $j=k-1$. The <primary ideal> condition applied to $r^ju\in\mathfrak q$ therefore forces $r^j\in\mathfrak q$, contrary to the choice of $k$.
Consequently every $x\in\mathfrak q$ lies in $(r^k)$, and
$$
\boxed{\mathfrak q=(r^k),\qquad k=\min\{n\geq1:r^n\in\mathfrak q\}.}
$$
Together with <1/e/i>, this classifies the <primary ideals for a regular principal prime>. The exponent is unique: if $(r^k)=(r^{k+1})$, then $r^k=r^{k+1}c$, and cancellation of the <non-zero-divisor> $r^k$ would give $1=rc$, contradicting that $(r)$ is proper. No <Noetherian> hypothesis is needed.
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