Quite a lot of algebraic number theory was invented through trying to prove Fermat's last theorem and other Diophantine problems.
For example, if I asked you to solve the equation $x^2 - y^2 = 5$ in integers it is very simple, you can factorise $(x+y)(x-y) = 5$ and solve the problem by linking to divisors of $5$, in order to get the solutions $x = \pm 3, y=\pm 2$.
Now say I ask you to solve the equation $y^3 = x^2 + 2$ in integers. This is not so easy if we work entirely in $\mathbb{Z}$ and use elementary methods. However, if we shift focus into a bigger ring of numbers, $\mathbb{Z}[\sqrt{-2}] = \{x+y\sqrt{-2}\,|\,x,y\in\mathbb{Z}\}$ then the problem again turns into a multiplicative problem:
$(x + \sqrt{-2})(x- \sqrt{-2}) = y^3$
so that solving the original Diophantine is really the same as solving a "product" style equation in $\mathbb{Z}[\sqrt{-2}]$.
The point of (basic) algebraic number theory is to study rings like this. How do the elements in these rings factorise?
In our problem above it turns out that the ring $\mathbb{Z}[\sqrt{-2}]$ has properties that are strangely close to properties of $\mathbb{Z}$. In fact the elements in this ring "factorise uniquely" into irreducible elements (the analogue of prime numbers in $\mathbb{Z}$).
The phrase "factorise uniquely" does not have quite the meaning you might think, we have to allow for multiplication of units (things that "divide $1$"). It is the ordering of $\mathbb{Z}$ allows us to consider unique factorisations into "positive primes".
There is also a notion of coprimality. This allows us to solve our problem since for odd $x$ it can be shown that $x\pm\sqrt{-2}$ are coprime in $\mathbb{Z}[\sqrt{-2}]$. But their product is a cube so (as in $\mathbb{Z}$), we must have that $x + \sqrt{-2} = (a+b\sqrt{-2})^3$ for some $a+b\sqrt{-2}\in \mathbb{Z}[\sqrt{-2}]$. Comparing coefficients lets you find the possibilities for $a,b$, hence for $x$.
The ideas of Lame and Kummer were to study FLT in the same way by considering the factorisation (for $\zeta$ a primitive $p$-th root of unity):
$z^p = x^p + y^p = (x + y)(x + \zeta y) ... (x + \zeta^{p-1} y)$
forming yet another product equation, now in the ring $\mathbb{Z}[\zeta]$.
Now this is not the entire story, since some of the rings we study in algebraic number theory do not have unique factorisation. For example the ring $\mathbb{Z}[\sqrt{-5}]$ does not since:
$6 = 2\times 3 = (1+\sqrt{-5})(1 - \sqrt{-5})$
gives two totally different factorisations of $6$. Actually the ring $\mathbb{Z}[\zeta]$ does not have unique factorisation for $p=23$, so that FLT could not be solved entirely by the above method.
The thing that stopped factorisation being unique was the fact that the ring wasn't big enough to factorise everything further into the same things.
Fortunately we can restore unique factorisation without having to extend! Through the genius of Kummer and Dedekind, they realised that by considering the "multiples" of an element as an object in its own right, we can reform factorisation in a way that becomes unique upto ordering.
In modern language these objects are called ideals of a ring. There is a notion of a prime ideal, capturing the notion of prime number. The different factorisations of $6$ above can be explained as reordering of the prime ideals in the factorisation of the ideal generated by $6$. These prime ideals are NOT generated by one element, so they dont correspond to "multiples" of something in $\mathbb{Z}[\sqrt{-5}]$, they correspond more to "multiples" of something that doesn't exist in the ring, but would exist after making an extension.
Kummer was able to prove a huge number of cases of FLT by using the ideal theory. This is outlined in many books.
Focus in algebraic number theory now turns to studying these algebraic constructions. We see that in a given "nice" ring, certain prime numbers may factorise, whereas others don't.
For example, in $\mathbb{Z}[i]$ we find that a prime $p$ factorises further if and only if $p=2$ or $p \equiv 1$ mod $4$. The factorisation of $2$ is different to the others in that $2 = (1+i)^2$ is not "square-free". All the others factorise into two different factors. We say that $2$ ramifies, primes $p\equiv 1$ mod $4$ split and primes $p\equiv 3$ mod $4$ are inert.
This congruence relationship describing the factorisation of primes is in some sense really explained by the values of the Legendre symbol $\left(\frac{-1}{p}\right)$, which is also explaining sums of two squares! Working in similar rings gives you the entire quadratic reciprocity law.
The goal of class field theory is to explain the splitting of primes in ANY extension of "number fields" to get similar characterisations in terms of congruences. In fact I just told a lie, we cannot yet do this for ANY extension, class field theory does it for abelian extensions (ones with abelian Galois group) but never-the-less it is quite a strong theory that has many applications (for example it solves the question of which primes can be written as $x^2 + ny^2$.
In the case of abelian extensions of $\mathbb{Q}$ we find that there are simple congruence conditions mod some integer $N$ that completely describe splitting behaviour of primes!
Another side of class field theory is the Cebotarev density theorem, which states essentially that most splitting types happen infinitely often. This is a huge generalisation of Dirichlet's theorem on primes in arithmetic progressions...in fact it provides an infinite amount of Dirichlet theorems, one for each abelian extension.
These days the (mostly unsolved) Langlands program is supposed to be filling in the gaps for non-abelian extensions but this is very difficult to understand and is not yet completely understood. When this is fully understood it will prove to be the holy grail of number theory, it will characterise in a huge way the splitting of primes.
Anyway, I hope this somewhat rushed introduction will whet your appetite. The book I first started with was Stewart/Tall - Algebraic number theory and fermat's last theorem. This is a good book to start you off. Also Lang - Algebraic number theory, Cox - Primes of the form $x^2 + ny^2$ and Childress - Class field theory are good ones to start with for class field theory.
Best Answer
Fermat's equation for cubes is a common introduction to lecture notes on algebraic number theory, because it motivates to study rings of integers in a number field, and partly has been developed even for such Diophantine problems, e.g., Kummer's work concerning generalizing factorization to ideals. For the equation $x^3+y^3=z^3$ the number field is $\mathbb{Q}(\zeta)$ with a third primitive root of unity $\zeta=e^{2\pi i/3}$. Its ring of integers is given by $\mathbb{Z}[\zeta]$, which is indeed a factorial ring (because it is Euclidean). Its units are given by $\pm 1,\pm \zeta,\pm\zeta^{-1}$. This is crucial to prove Euler's result:
Theorem(Euler $1770$): The equation $x^3+y^3=z^3$ has no non-trivial integer solutions.
The proof uses divisibility properties of the ring $\mathbb{Z}[\zeta]$, starting from the equation $$ z^3=x^3+y^3=(x+y)(x+\zeta y)(x+\zeta^2y). $$ The first case is $p=3\nmid xyz$. We may suppose that $x,y,z$ are coprime. We have $z^3\equiv \pm 1\bmod 9$ and $x^3+y^3\equiv -2,0,2 \bmod 9$, so that $x^3+y^3\neq z^3$, a contradiction. Hence we suppose that $3\mid xyz$, i.e., say, $3\mid z$ and $3\nmid xy$. Now we reformulate the equation as $$ x^3+y^3=(3^mz)^3, $$ with $x,y,z$ pairwise coprime and $3\nmid xyz$, where we have solved the case $m=0$. The idea is now to use descent, i.e., to reduce it to the case $m=0$. The above equation becomes $$ (3^mz)^3=(x+y)(x+\zeta y)(x+\zeta^2y), $$ where the three factors are not coprime, because $1-\zeta$ is a common factor, because of $3=(1-\zeta)(1-\zeta^2)$, so that $(1-\zeta)\mid 3\mid (x+y)$. However, since $\mathbb{Z}[\zeta]$ is factorial, all factors are cubes, i.e. , $x+y=3^{3m-1}c^3$ with some $c\in \mathbb{Z}$, and so on. This finishes the proof, after some computations in this ring.
Unfortunately, this idea does not work for $x^p+y^p=z^p$ for primes $p$, except for $p\le 19$, because otherwise the ring of integers $\mathbb{Z}[\zeta_p]$ is no longer factorial.