Let $\mathcal{O}_K$ be the ring of integers of a quadratic number field $K$. If $\mathfrak{q}$ is a prime ideal of $\mathcal{O}_K$, then $N(\mathfrak{q})=\mathfrak{q}\overline{\mathfrak{q}}\cap \mathbb{Q}=p^n\mathbb{Z}$ for some prime power $p^n$, $n=1,2$.
Let $(p)=\mathfrak{a}\mathfrak{b}$ by the ideal decomposition of $(p)$ in $\mathcal{O}_K$. If $(p)$ splits then $\mathfrak{b}=\overline{\mathfrak{a}}$ is prime, if $(p)$ is inert then $\mathfrak{a}=(p)$ is prime and $\mathfrak{b}=(1)$, and if $(p)$ ramifies then $\mathfrak{b}=\mathfrak{a}$ is prime.
Then we have that $p^n\mathcal{O}_K=\mathfrak{a}^n\mathfrak{b}^n$, but then $\mathfrak{q}=\mathfrak{a}$ or $\mathfrak{q}=\mathfrak{b}$ because $\mathfrak{q}\overline{\mathfrak{q}}=p^n\mathcal{O}_K$.
If $(p)$ splits, then $\mathfrak{q}\overline{\mathfrak{q}}=\mathfrak{a}\mathfrak{b}=(p)$.
If $(p)$ is inert, then $\mathfrak{q}\overline{\mathfrak{q}}=(p^2)$.
If $(p)$ ramifies, then $\mathfrak{q}\overline{\mathfrak{q}}=\mathfrak{a}\overline{\mathfrak{a}}=(p)$ (because $\mathfrak{a}^2\overline{\mathfrak{a}}^2=\mathfrak{a}^4=(p^2)$).
Question: Is this analysis correct? That is, are the only prime ideals $\mathfrak{q}$ in a quadratic number field such that $N(\mathfrak{q})=p^2$ the inert prime ideals $(p)$? Is there an easier way to show this?