A trick I have seen several times: If you want to show that some rational number is an integer (i. e., a divisibility), show that it is an algebraic integer. Technically, it is then an application of commutative algebra (the integral closedness of $\mathbb Z$, together with the properties of integral closure such as: the sum of two algebraic integers is an algebraic integer again), but since you define algebraic number theory as the theory of algebraic numbers, you may be interested in this kind of applications.
Example: Let $p$ be a prime such that $p\neq 2$. Prove that the $p$-th Fibonacci number $F_p$ satisfies $F_p\equiv 5^{\left(p-1\right)/2}\mod p$.
Proof: We can do the $p=5$ case by hand, so let us assume that $p\neq 5$ for now. Then, $p$ is coprime to $5$ in $\mathbb Z$. Let $a=\frac{1+\sqrt5}{2}$ and $b=\frac{1-\sqrt5}{2}$. The Binet formula yields $F_p=\displaystyle\frac{a^p-b^p}{\sqrt5}$. Now, $a^p-b^p\equiv\left(a-b\right)^p\mod p\mathbb Z\left[a,b\right]$ (by the idiot's binomial formula, since $p$ is an odd prime). Note that $p$ is coprime to $5$ in the ring $p\mathbb Z\left[a,b\right]$ (since $p$ is coprime to $5$ in the ring $\mathbb Z$, and thus there exist integers $a$ and $b$ such that $pa+5b=1$). Now,
$\displaystyle F_p=\frac{a^p-b^p}{\sqrt5}\equiv\frac{\left(a-b\right)^p}{\sqrt5}$ (since $a^p-b^p\equiv\left(a-b\right)^p\mod p\mathbb Z\left[a,b\right]$ and since we can divide congruences modulo $p\mathbb Z\left[a,b\right]$ by $\sqrt5$, because $p$ is coprime to $5$ in $p\mathbb Z\left[a,b\right]$)
$\displaystyle =\frac{\left(\sqrt5\right)^p}{\sqrt5}$ (since $a-b=\sqrt5$)
$=5^{\left(p-1\right)/2}\mod p\mathbb Z\left[a,b\right]$.
In other words, the number $F_p-5^{\left(p-1\right)/2}$ is divisible by $p$ in the ring $\mathbb Z\left[a,b\right]$. Hence, $\frac{F_p-5^{\left(p-1\right)/2}}{p}$ is an algebraic integer. But it is also a rational number. Thus, it is an integer, so that $p\mid F_p-5^{\left(p-1\right)/2}$ and thus $F_p\equiv 5^{\left(p-1\right)/2}\mod p$, qed.
Yes, there exist purely algebraic conditions on a Dedekind domain which hold for all rings of integers in global fields and which imply that the class group is finite.
For a finite quotient domain $A$ (i.e., all non-trivial quotients are finite rings), a non-zero ideal $I\subseteq A$ and a non-zero $x\in A$, let $N_{A}(I)=|A/I|$ and $N_{A}(x)=|A/xA|$. Also define $N_{A}(0)=0$.
Call a principal ideal domain $A$ a basic PID if the following conditions are satisfied:
$A$ is a finite quotient domain,
for each $m\in\mathbb{N}$,$$\#\{x\in A\mid N_{A}(x)\leq m\}>m$$
(i.e., $A$ has “enough elements of small norm”),
there exists a constant $C\in\mathbb{N}$ such that for all $x,y\in A$,
$$N_{A}(x+y)\leq C\cdot(N_{A}(x)+N_{A}(y))$$
(i.e., $N_{A}$ satisfies the “quasi-triangle inequality”).
Theorem. Let $A$ be a basic PID and let $B$ be a Dedekind domain which is finitely generated and free as an $A$-module. Then $B$ has finite ideal class group.
For the proof, see here.
It is easy to verify that $\mathbb{Z}$ and $\mathbb{F}_q[t]$ are basic PIDs, so the ring of integers in any global field satisfies the hypotheses of the above theorem (using the non-trivial fact that rings of integers in global fields are finitely generated over one of these PIDs).
More generally, one can take the class of overrings of Dedekind domains which are finitely generated and free over some basic PIDs. Since it is known that an overring of a Dedekind domain with finite class group also has finite class group, this gives a wider class of algebraically defined Dedekind domains (including $S$-integers like $\mathbb{Z}[\frac{1}{p}]$) with finite class group.
Added: The second condition for basic PIDs can be relaxed to: there exists a constant $c\in\mathbb{N}$ such that for each $m\in\mathbb{N}$,
$$
\#\{x\in A\mid N_{A}(x)\leq c\cdot m\}\geq m.
$$
Best Answer
For a course on algebraic number theory, you certainly can prove the finiteness of the class group without Minkowski's theorem. For example, if you look in Ireland-Rosen's book you will find a proof there which they attribute to Hurwitz. It gives a worse constant (which depends on a choice of $\mathbf Z$-basis for the ring of integers of the number field; changing the basis can shrink the constant, but it's still generally worse than Minkowski's) but it is computable and you can use it to show, say, that $\mathbf Z[\sqrt{-5}]$ has class number 2.
As for the history of the proof of the unit theorem, it was proved by Dirichlet using the pigeonhole principle. If you think about it, Minkowski's convex body theorem is a kind of pigeonhole principle (covering the convex body by translates of a fundamental domain for the lattice and look for an overlap). You can find a proof along these lines in Koch's book on algebraic number theory, published by the AMS. Incidentally, Dirichlet himself proved the unit theorem for rings of the form $\mathbf Z[\alpha]$; the unit theorem is true for orders as much as for the full ring of integers (think about Pell's equation $x^2 - dy^2 = 1$ and the ring $\mathbf Z[\sqrt{d}]$, which need not be the integers of $\mathbf Q(\sqrt{d})$), even though some books only focus on the case of a full ring of integers. Dirichlet didn't have the general conception of a full ring of integers.
One result which Minkowski was able to prove with his convex body theorem that had not previously been resolved by other techniques was Kronecker's conjecture (based on the analogy between number fields and Riemann surfaces, with $\mathbf Q$ being like the projective line over $\mathbf C$) that every number field other than $\mathbf Q$ is ramified at some prime.