Things become more difficult for the ##p=3## case. The basic idea is to translate the additive problem
$$
a^p+b^p=c^p
$$
into a multiplicative problem by writing it as
$$
c^p=a^p+b^p=(a+b)\cdot(a+\zeta b)\cdot (a+ \zeta^2 b)\cdot \ldots \cdot (a+\zeta^{p-1}y)
$$
where ##\zeta## is a root of ##x^p-1=0.## This is a complex number. The crucial point, however, is that it is a polynomial equation with coefficients in the ring ##\mathbb{Z}[\zeta]## as
@martinbn has mentioned in post #7. Now we have a multiplicative equation in a ring. This ring has primes. A prime is a number that, if it divides a product, then it has to divide one of its factors. E.g. ##6## divides ##3\cdot 4## but it doesn't divide ##3## or ##4,## so it is not prime. On the other hand, ##3## divides ##2\cdot 6## so it has to divide either ##2## which it does not, or ##6,## which it does divide. And this is true for any factorization of ##1\cdot 12=2\cdot 6= 3\cdot 4,## i.e. ##3## is prime. Hence, if a prime element of ##\mathbb{Z}[\zeta]## divides ##c^p## then it has to divide one of the factors ##a+\zeta^kb## for some ##k<p.##
This is the basic idea.
In the case ##p=3## we may set ##\zeta=\dfrac{1}{2}+ i\dfrac{\sqrt{3}}{2}.## Then ##
\zeta^2=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}## and ##\zeta^3=-1\, , \,\zeta^2-\zeta=-1.## Now,
$$
(a+b)(a-b\zeta)(a+b\zeta^2)=(a+b)\left[a^2-ab+b^2\right]=c^3.
$$
##\mathbb{Z}[\zeta]## has unique prime factor decompositions and is a Euclidean integral domain. Its elements all can be written as ##x+y\zeta +z\zeta^2## with integers ##x,y,z.## However, all numbers ##m^2+3n^2## are reducible, i.e. ##3## is not prime. Its units are all sixth roots of unity, i.e., solutions to ##x^6=1.## All positive integer primes ##p\equiv 2\pmod{3}## are prime in ##\mathbb{Z}[\zeta],## too. All positive integer primes ##p\equiv 1\pmod{3}## are a product of two complex conjugate primes in ##\mathbb{Z}[\zeta].## All those, ##1-\omega^2=2-\omega,## and their associates are the prime elements of ##\mathbb{Z}[\zeta].## Associate means up to units. Units are those elements that can be inverted. Of course, we can always multiply units to a number without changing its primality. ##-3## is also a prime integer, however, not in ##\mathbb{Z}[\zeta]## anymore. Anyway, this ring has all the nice properties I listed and allows us to consider divisibilities. These properties are not obvious, so either you trust me or you have to prove them.
The point is, that with primes, a unique prime factor decomposition, and a Euclidean algorithm, a norm function ##N(a+y \sqrt{-3})=x^2+3y^3## to distinguish between larger and smaller numbers, we have all ingredients for number theory on ##\mathbb{Z}[\zeta]## and the investigation of the prime factors of ##c^3## in the now multiplicative equation.
##\mathbb{Z}[\zeta]## are called Eisenstein numbers in case you want to look it up. Also, I think the first source I listed (D. Zachow) has more about Eisenstein numbers and the rings ##\mathbb{Z}[\zeta]## and a proof for the case ##p=3,## but it is in German.
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