For fixed $n \in \mathbb{N}$ consider integer solutions to
$$x^3+y^3+z^3=n \qquad (1) $$
If $n$ is a cube or twice a cube, identities exist.
Elkies suggests no other polynomial identities are known.
For which $n$ (1) has infinitely many integer solutions?
Added
Is there $n$, not a cube or twice a cube, which allows infinitely
many solutions?
Added 2019-09-23:
The number of solutions can be unbounded.
For integers $n_0,A,B$ set $z=Ax+By$ and consider
$x^3+y^3+(Ax+By)^3=n_0$. This is elliptic curve
and it may have infinitely many rational points
coming from the group law. Take $k$ rational points
$(X_i/Z_i,Y_i/Z_i)$. Set $Z=\rm{lcm}\{Z_i\}$.
Then $n_0 Z^3$ has the $k$ integer solutions $(Z X_i/Z_i,Z Y_i/Z_i)$.
Best Answer
For $n\equiv \pm 4\pmod{9}$ there is no solution to $(1)$. Otherwise, for $n\ge 1$, it is conjectured that there are always solutions, even infinitely many. There are no analytic results, but heuristics suggest that given $n$, not $0$ or $\pm 4\pmod{9}$, solutions should occur infinitely often, asymptotically $c\log(N)$ solutions in $|x|,|y|,|z|<N$, see papers of Conn and Vaserstein.
The topic has been discussed quite frequently, see also sum of three cubes and parametric solutions, Are nontrivial integer solutions known for $x^3+y^3+z^3=3$?, Efficient computation of integer representation as a sum of three squares, etc. For a collection on polynomial parametric solutions, see https://sites.google.com/site/tpiezas/010.
See also the Euler-Binet solutions to $x^3+y^3=z^3+w^3$, $$ x = 1 − (p − 3q)(p^2 + 3q^2), $$ $$ y = −1 + (p + 3q)(p^2 + 3q^2), $$ $$ z = (p + 3q) − (p^2 + 3q^2)^2, $$ $$ w = −(p − 3q) + (p^2 + 3q^2)^2. $$