= Solution
The <ring of integers> is $\mathcal O_k=\mathbb Z[\sqrt7]$, since $7\equiv3\pmod4$, and its <discriminant> is $28$. Rationalizing gives
$$
\varepsilon=8+3\sqrt7,\qquad\varepsilon^{-1}=8-3\sqrt7,\qquad N(\varepsilon)=1.
$$
Both are algebraic integers, so \b[$\varepsilon$ is a <unit>]. Set $\mathfrak p=(3+\sqrt7)$. Since $(3+\sqrt7)^2=2\varepsilon$,
$$
\boxed{\mathfrak p^2=(2).}
$$
Also $N(\mathfrak p)=|N(3+\sqrt7)|=2$. It is the unique ideal of <ideal norm> two: any quotient of size two is $\mathbb F_2$, and the image of $\sqrt7$ must satisfy $z^2=1$, hence be one. The <kernel of a ring homomorphism> onto $\mathbb F_2$ is uniquely $(2,1+\sqrt7)$, which is $\mathfrak p$.
The given Minkowski constant means every <ideal class> has an integral representative of <ideal norm> at most $\frac12\sqrt{28}=\sqrt7<3$. Such a <ideal norm> is one or two. The norm-one ideal is the whole ring, and the sole norm-two ideal is principal. Thus \b[the <class number> is one].
If $x^2-7y^2=2$, the <principal ideal> $(x+y\sqrt7)$ has <ideal norm> two and hence equals $\mathfrak p$. Its generator differs from $3+\sqrt7$ by a <unit>. Conversely multiplication by a norm-one <unit> preserves the <field norm> two. With the assumed <fundamental unit>, all <units> are $\pm\varepsilon^n$; in fact a norm-minus-one <unit> is impossible since $x^2\equiv-1\pmod7$ has no solution. Therefore the complete answer is
$$
\boxed{x+y\sqrt7=\pm(8+3\sqrt7)^n(3+\sqrt7),\qquad n\in\mathbb Z.}
$$
For $n=1$ and the positive sign, the product is $45+17\sqrt7$, giving \b[$(x,y)=(45,17)$]; $45^2-7\cdot17^2=2$ checks the result.
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