Diophantine equations solved using algebraic numbers?

Perhaps the simplest example is the parametrization of primitive Pythagorean triples $ z^2 = x^2 + y^2 = (x-y\,i)(x+y\,i).\,$ The key idea behind the proof is this: $\ x-y\,i,\ x+y\,i\ $ are coprime factors of a square in a UFD, hence they must themselves be squares (up to unit factors). So $\ x + y\, i = (m + n\ i)^2 =\ m^2 - n^2 + 2mn\, i,\,$ which is the well-known parametrization.

Similarly we can solve low degree cases of Fermat's Last Theorem by employing analogous factorizations over certain rings of algebraic integers. For example, Gauss showed there are no solutions for exponent 3 by working in the ring of integers of $\ \mathbb Q(\sqrt{-3}),\,$ and Dirichlet did similarly for exponent 5 using $\ \mathbb Q(\sqrt{5})\,.$ Later Kummer generalized these techniques to handle all regular prime exponents by working over rings of cyclotomic integers. For a nice exposition see Ribenboim: 13 lectures on Fermat's last theorem. Weil nicely summarizes the essence of these techniques in his Number Theory, Ch.IV,S.VI,p.335:

The classic example is certain Mordell curves. I can never quite remember which ones have this property off the top of my head, but for example I think $y^2 = x^3 - 1$ is one of them. (Write it as $y^2 + 1 = x^3$ and factor the LHS in $\mathbb{Z}[i]$.) There might be a list in some books on Diophantine equations and/or algebraic number theory and/or elliptic curves.

There is also the regular case of Fermat's last theorem. I don't know if this counts as "easy," though.

Euler famously used algebraic integers to solve diophantine problems, though not always exactly right.

For example, in article 182 of Chapter XII, Part II, of his Elements of Algebra (p. 396 in my brand new Cambridge Library Collection edition), he writes:

Let, therefore, the formula $x^2+cy^2$ be proposed, and let it be required to make it a square. As it is composed of the factors $(x+y\sqrt{-c})(x-y\sqrt{-c})$, these factors must either be squares, or squares multiplied by the same number. For, if the product of two numbers, for example, $pq$, must be a square, we must have $p=r^2$ and $q=s^2$; that is to say, each factor is of itself a square; or $p=mr^2$ and $q=ms^2$; and therefore these factors are squares multiplied by the same number. For which reason, let us make $x+y\sqrt{-c} = m(p+q\sqrt{-c})^2$; it will follow that $x-y\sqrt{-c} = m(p-q\sqrt{-c})^2$, and we shall have $x^2+cy^2 = m^2(p^2+cq^2)^2$, which is a square. Farther, in order to determine $x$ and $y$, we have the equations $x+y\sqrt{-c} = mp^2 + 2mpq\sqrt{-c} -mcq^2$, and $x-y\sqrt{-c} = mp^2-2mpq\sqrt{-c} - mcq^2$; in which $x$ is necessarily equal to the rational part, and $y\sqrt{-c}$ to the irrational part; so that $x=mp^2 - mcq^2$ and $y\sqrt{-c} = 2mpq\sqrt{-c}$, or $y=2mpq$; and these are the values of $x$ and $y$ that will transform the expression $x^2+cy^2$ into a square, $m^2(p^2+cq^2)^2$, the root of which is $mp^2+mcq^2$.

(Note that Euler in fact finds sufficient conditions, but not necessary ones, here, and there is an implicit assumption of unique factorization). In Article 187, he proceeds along the same lines to discuss when $ax^2+cy^2$ will be a cube, working with the factorization $ax^2+cy^2 = (x\sqrt{a}+y\sqrt{-c})(x\sqrt{a} - y\sqrt{-c})$.

Euler himself knew this did not quite work out in general: in Article 195 he discusses when $2x^2-5$ is a cube; his method yields the answer that it is never a cube, but yet he notes that $x=4$ gives $2x^2 - 5 = 27 = 3^3$; his "investigation" into the failure involves some rather strenuous hand-waving, concluding that the problem is that we have a difference instead of a sum as before.