The above proposition is a special case of the following proposition.
Proposition Let $f(X)$ be a monic irreducible polynomial of degree 3 in $\mathbb{Z}[X]$. Let $d$ be the discriminant of $f(X)$. Let $K = \mathbb{Q}(\sqrt{d})$. Let $L$ be the splitting field of $f(X)$ over $\mathbb{Q}$. Suppose the following conditions hold.
(1) $|d|$ is a prime number.
(2) The class number of $K$ is 3.
(3) $f(X) \equiv (X - s)^2(X - t)$ (mod $d$), where $s$ and $t$ are rational integers and $s$ and $t$ are distinct mod $d$.
Then $L$ is the Hilbert class field of $K$.
Proof: Since the degree of $f(X)$ is 3, $[L : \mathbb{Q}] \leq 6$. Since $f(X)$ is irreducible, $3|[L : \mathbb{Q}]$. By (1), $[K : \mathbb{Q}] = 2$. Hence $2|[L : \mathbb{Q}]$. Hence $[L : \mathbb{Q}] = 6$. Hence $[L : K] = 3$. Therefore, by (2), it suffices to prove that every prime ideal of $K$ is unramified in $L$.
Let $Q$ be a prime ideal of $K$ lying over a prime number $q \neq p$, where $p = |d|$. By the application of the proposition of this question, $q$ is unramified in $L$. Hence $Q$ is unramified in $L$.
Let $P$ be a prime ideal of $K$ lying over $p$. It remains to prove that $P$ is unramified in $L$.
Let $\theta$ be a root of $f(X)$ in $L$. Let $M = \mathbb{Q}(\theta)$. Since $f(X)$ is irreducible, $[M : \mathbb{Q}] = 3$. Hence $[L : M] = 2$.
We denote by $\mathcal{O}_K, \mathcal{O}_M, \mathcal{O}_L$ the rings of integers in $K$, $M$, $L$ respectively.
Let $D_M$ be the discriminant of $M$. It is well known that $d = k^2 D_M$ for some rational integer $k$. Since $k^2 = 1$ by (1), $d = D_M$. Hence $\mathcal{O}_M = \mathbb{Z}[\theta]$. It is is well known(e.g. Milne's online course note) that $p\mathcal{O}_M = \mathfrak{p}^2\mathfrak{q}$ by (3), where $\mathfrak{p}$ and $\mathfrak{q}$ are distinct prime ideals of $\mathcal{O}_M$.
Since $[L : M] = 2$, We have the following patterns of the prime decompositions in $L$.
(1) $\mathfrak{p}\mathcal{O}_L = \mathfrak{P}$.
(2) $\mathfrak{p}\mathcal{O}_L = \mathfrak{P_1}\mathfrak{P_2}$, where $\mathfrak{P_1} \neq \mathfrak{P_2}$.
(3) $\mathfrak{p}\mathcal{O}_L = \mathfrak{P}^2$.
(1)' $\mathfrak{q}\mathcal{O}_L = \mathfrak{Q}$.
(2)' $\mathfrak{q}\mathcal{O}_L = \mathfrak{Q_1}\mathfrak{Q_2}$, where $\mathfrak{Q_1} \neq \mathfrak{Q_2}$.
(3)' $\mathfrak{q}\mathcal{O}_L = \mathfrak{Q}^2$.
Since $L/\mathbb{Q}$ is Galois, each ramification index of prime ideals of $L$ lying over $p$ is the same. Hence only the combination of (2) and (3)' is possible. Hence $p\mathfrak{O}_L = \mathfrak{P_1}^2\mathfrak{P_2}^2\mathfrak{Q}^2$.
Since $p$ ramifies in $K$, $p\mathfrak{O}_K = P^2$. Hence by the above result, $P$ is unramified in $L$. QED