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.