What heuristic suggest for the number of solutions of $x^n+y^n=A$? For integers $x,y,n,A$ with $n>1$ and $A>0$ we are interested
how many solutions $x^n+y^n=A$ has for fixed $n$ and infinitely many
$A$.
What is unconditionally known $n=2$ or $n=3$ the number of solutions
is unbounded. Take many rational solutions to $x^n+y^n=A_0$
with the group law. Scale to integers by multiplying by the lcm
of the denominators.

Q1 for $n=3$, without using the group law, are there arguments
for unboundedness of solutions?


Q2 For $n=3$, without using the group law, can we find many,
say 100 solutions? This is related to the rank of the elliptic curve.

For $n>4$, it is conjectured that up to automorphism
there are no solutions to $x_1^n+y_1^n=x_2^n+y_2^n=A$.

Q3 Are there arguments for the above conjecture?

 A: (1) You don't need the full group law, of course. Start with a rational solution to $x^3+y^3=A_0$, use the tangent line to find a new solution, use the tangent line to that solution to find another solution, etc. (I think this idea may be due to Bachet in the 1600s. In any case, it's simple algebra.) Then clear denominators. But maybe you're asking for an argument that doesn't use analytic geometry?
(2) A much more interesting question, I think, is whether you can find $A$ having a large number of integer solutions satisfying $\gcd(x,y)=1$. That's what's really related to the rank, in the sense that there's a proven inequality
$$ \#\{(x,y)\in\mathbb Z^2 : x^3+y^3=A,\;\gcd(x,y)=1\} \le C^{1+\text{rank }E_A(\mathbb Q)}, $$
where the constant $C$ is an absolute constant, and where $E_A$ is the elliptic curve $x^3+y^3=A$.
A: There is a heuristic, which doesn't use the group law, that there should be infinitely many relatively prime integer solutions for $x^3 +y^3 = A z^3$, and therefore unboundedness by multiplying $A$ by the cube of the lcm of the $z$s.
The main problem with this heuristic is it predicts infinitely many triples $(x,y,z)$ for all $A$, but we know there is infiniteness for only some $A$.
The heuristic is the very simple randomness-based heuristic. If we restrict attention to $x,y,z$ with $2^n <\max (|x|,|y|,|z|) < 2^{n+1}$, there are about $ C \cdot 2^{3n}$ relatively prime triples in that range, and then if we imagine each has a probability at least $1 / (2 \cdot (2 + A) \cdot (2^{n+1})^3)$ of being a solution, the total number of solutions in that range should be about $C/ (2^4 \cdot (2+A))$.
Summing over all $n$, we get $\infty$.
This heuristic, with a modification for local factors, is the one used by Heath-Brown to study sums of three cubes and other inhomogeneous cubics, where there is no group law. So, if you're bothered by the fact that this heuristic seems to predict the wrong answer, asking for a heuristic avoiding the group law might be a bit tricky, because the group law is exactly the reason that simple heuristics fail.
One can use this type of heuristic to check that $x^n + y^n  = z^n+w^n$ should have very few solutions, with solutions getting less likely as $|x|,|y|,|z|,|w|$ grow or $n$ grows. This, combined with a computer check for no small solutions, is presumably a large part of the reason that some have conjectured no nontrivial solutions there.
A: I think the conjecture that for $n \geq 5$ and for all integers $k \geq 1$,
$$\displaystyle x^n + y^n = k$$
has at most two positive integer solutions is much older, but a modern heuristic comes from the Bombieri-Lang conjecture. Indeed, I believe it is known how to find all curves of small degree on a Fermat surface $x^n + y^n = u^n + v^n$ for $n$ sufficiently large (though I am not sure if $n$ can be taken as low as $5$); see for example this paper of Browning and Heath-Brown. In any case it is known that any smooth algebraic surface of degree $n$ contains $O_n(1)$ curves of degree at most $n-2$; this is a result of Colliot-Thelene (see the appendix of this paper). It is then expected, given Bombieri-Lang, that these low-degree curves contain all of the rational points. Moreover, using explicit techniques which are available for Fermat surfaces we can check that none of these curves except for the obvious lines are rational. This gives the heuristic that all solutions to $x^n + y^n = u^n + v^n$, i.e., repeated solutions to $x^n + y^n = A$ for any integer $A$, are accounted for by the lines. Heath-Brown shows in the above linked paper that such lines correspond to automorphisms of the binary form $x^n + y^n$, which then gives the conjecture.
Of course, we don't know how to prove Bombieri-Lang in this case (or really any case beyond subvarieties of abelian varieties as far as I know), so this remains merely a heuristic.
