TITLE: Diophantine equations solved using algebraic numbers?
QUESTION [9 upvotes]: On Mathworld it says some diophantine equations can be solved using algebraic numbers. I know one example which is factor $x^2 + y^2$ in $\mathbb{Z}[\sqrt{-1}]$ to find the Pythagorean triples.
I would be very interested in finding some examples of harder equations (not quadratic) which are also solved easily using algebraic numbers. Thank you!

REPLY [3 votes]: An interesting example which appears in section $4.9$ of Stewart and Tall's Algebraic Number Theory and Fermat's Last Theorem is the solution of the Ramanujan-Nagell equation given by
$$x^2 + 7 = 2^n$$
The idea is to work over the quadratic field $\mathbb{Q}(\sqrt{-7})$ and use the fact that it has unique factorization in order to compare two factorizations at the core of the argument. 
The argument needs to take care of different cases separately and does not run as "smoothly" as the arguments in some of the examples mentioned in the other answers, but that's probably in some sense due to the fact that this equation involves an exponential, which only makes the already difficult problem of solving a Diophantine equation worse.