= Solution
The <Minkowski convex body theorem> states that if $\Lambda\subset\mathbb R^n$ is a full <lattice> and $C$ is a measurable, convex, centrally symmetric set with
$$
\operatorname{vol}(C)>2^n\operatorname{covol}(\Lambda),
$$
then $C$ contains a nonzero point of $\Lambda$.
Here $d\not\equiv3\pmod4$, so the <ring of integers of a quadratic field> is
$$
\mathcal O_K=\mathbb Z[\sqrt{-d}].
$$
Under its complex embedding it has real-lattice covolume $\sqrt d$. An integral ideal $I$ has index $N(I)$ in $\mathcal O_K$, so its covolume is $\sqrt d\,N(I)$. Apply Minkowski's theorem to the disk $|z|\le R$, whose area is $\pi R^2$. Taking $R^2$ down to $4\sqrt d\,N(I)/\pi$ gives a nonzero $\alpha\in I$ with
$$
|\alpha|^2\le\frac{4\sqrt d}{\pi}N(I).
$$
Since the <field norm> is $N_{K/\mathbb Q}(\alpha)=|\alpha|^2$ in this imaginary quadratic field, this proves the <imaginary quadratic Minkowski disk bound>
$$
\boxed{0<|N_{K/\mathbb Q}(\alpha)|
\le\frac{4\sqrt d}{\pi}N(I)}.
$$
Because $(\alpha)\subseteq I$, the ideal
$$
J=(\alpha)I^{-1}
$$
is integral, represents the inverse class of $I$, and has
$$
N(J)=\frac{|N_{K/\mathbb Q}(\alpha)|}{N(I)}
\le\frac{4\sqrt d}{\pi}.
$$
Thus every ideal class has a <bounded-norm ideal representatives>[bounded-norm representative]. Only finitely many integral ideals have bounded norm, proving the <finiteness of the ideal class group>.
Now take $K=\mathbb Q(\sqrt{-22})$. Its ring of integers is $\mathbb Z[\sqrt{-22}]$, its discriminant is $-88$, and
$$
\frac{4\sqrt{22}}{\pi}<6.
$$
Every class therefore has an integral representative of norm at most five. The polynomial $X^2+22$ is irreducible modulo three and modulo five, so both primes are inert and there are no ideals of norm three or five. At two it reduces to $X^2$, and the <splitting of rational primes in a quadratic field> gives
$$
(2)=\mathfrak p_2^2,
\qquad
\mathfrak p_2=(2,\sqrt{-22}),
\qquad
N(\mathfrak p_2)=2.
$$
An ideal of norm four is $\mathfrak p_2^2=(2)$ and is principal. The ideal $\mathfrak p_2$ is not principal, since a generator would have norm two, whereas
$$
a^2+22b^2=2
$$
has no integer solution. Consequently the <Ideal class group of Q of square root of minus twenty-two> is
$$
\boxed{\operatorname{Cl}(\mathbb Q(\sqrt{-22}))\cong C_2}.
$$
Suppose finally that $y^3=x^2+22$. If $x$ were even, the right-hand side would be $2$ modulo four, which is not a cube modulo four. Hence $x$ and $y$ are odd. In $\mathcal O_K$,
$$
(x+\sqrt{-22})(x-\sqrt{-22})=(y)^3.
$$
The two ideals on the left are coprime. Indeed, a common prime ideal would divide both $2x$ and $2\sqrt{-22}$. It cannot lie above two because their norms are odd. It would therefore lie above eleven and force $11\mid x$; but then $x^2+22$ has $11$-adic valuation exactly one, impossible for the cube $y^3$.
By <unique factorization of ideals in a number field>, coprimality implies
$$
(x+\sqrt{-22})=\mathfrak a^3
$$
for some ideal $\mathfrak a$. The class group has order two, so it has no element of order three; $[\mathfrak a]^3=1$ therefore forces $\mathfrak a$ to be principal. The only units are $\pm1$, both of which can be absorbed into a cube, and hence
$$
x+\sqrt{-22}=(a+b\sqrt{-22})^3
$$
for integers $a,b$. Comparing the coefficient of $\sqrt{-22}$ gives
$$
b(3a^2-22b^2)=1.
$$
Thus $b=\pm1$. For $b=1$ this requires $3a^2=23$, and for $b=-1$ it requires $3a^2=21$; neither is possible. Therefore the result on <no integral solutions to y cubed equals x squared plus twenty-two> is
$$
\boxed{y^3=x^2+22\text{ has no integer solutions}}.
$$
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