This is a consequence of the well-known fact that $\Bbb Z[\sqrt{-2}]$ is a Euclidean domain. Viewing the ring as a lattice inside of $\Bbb C$ lets you make a geometric argument that for any $a,b\in\Bbb Z[\sqrt{-2}]$ with $b\neq 0$, there exist $q,r\in\Bbb Z[\sqrt{-2}]$ such that $a = bq + r$ with $N(r) = r\cdot\overline{r} < N(b)$. (See here for a proof.)
Once you know this, the claim follows from applying the division algorithm with $b = \alpha$. An arbitrary element of $\Bbb Z[\sqrt{-2}]/(\alpha)$ is of the form $a + (\alpha)$, where $a\in\Bbb Z[\sqrt{-2}]$. But the division algorithm tells you that there exists $q,r\in\Bbb Z[\sqrt{-2}]$ such that $a = \alpha\cdot q + r$, with $N(r) < N(\alpha) = 20^2 + 2\cdot 5^2$. Then $a - r = \alpha\cdot q\in (\alpha)$, so that $a + (\alpha) = r + (\alpha)$ as elements of $\Bbb Z[\sqrt{-2}]/(\alpha)$. Thus, an arbitrary element of the quotient has a representative with norm less than $\alpha$.
You can also see this result using results one would see in a first course in algebraic number theory. $\Bbb Z[\sqrt{-2}]$ is the ring of integers of the number field $\Bbb Q(\sqrt{-2})$. The ring of integers of a number field is a Dedekind domain, and every nonzero ideal in such a ring factors uniquely into a product of prime ideals. Then if $(\alpha) = \prod_i\mathfrak{p}_i^{e_i}$, where the $\mathfrak{p}_i$ are prime ideals, you have
$$
\Bbb Z[\sqrt{-2}]/(\alpha)\cong \prod_i \Bbb Z[\sqrt{-2}]/\mathfrak{p}_i^{e_i}
$$
by the Chinese remainder theorem. Each of the factors in this product of quotients is in fact a finite dimensional vector space over $\Bbb F_p$ for some prime $p$, and hence the ring must be finite.
One last approach: once you decompose $(\alpha)$ into a product of primes, in a general Dedekind domain $\mathfrak{a}\mid\mathfrak{b}$ if and only if $\mathfrak{a}\supseteq\mathfrak{b}$ (where $\mathfrak{a}$ and $\mathfrak{b}$ are ideals in the Dedekind domain). You could use this fact coupled with the bijection between ideals of $R/I$ and ideals of $R$ containing $I$ that exists for any ring $R$ and ideal $I\subseteq R$. There are only finitely many ideals $I\subseteq\Bbb Z[\sqrt{-2}]$ containing $(\alpha)$, because $I\mid(\alpha)$ if and only if $I$ factors into prime ideals as $\prod_i\mathfrak{p}_i^{e'_i}$ with $e'_i\leq e_i$ for all $i$.
