What are the characteristics of the values of $c$ for which the equation $x^3 + y^3 = cz^3$ has pairwise coprime non-zero integral solutions where $x \neq \pm y$ ? For instance, it is known that $c$ cannot be a power of 2.
A few computational searches seems to suggest that $c$ must be a prime power. E.g., for $c=9$, the equation defines an elliptic curve with the point $(1, 2, 1)$, from which infinitely many coprime solutions can be derived.
You don't say that $c\in\mathbb Z$, but you imply that it is. Since we can absorb cube factors of $c$ into $z$, we may assume that $c$ is cube-free. Then your assertion that $c$ cannot be a power of $2$ is simply the assertion that $c$ cannot be 1, 2, or 4. In general, let $E_c$ be the elliptic curve $$ E_c : x^3 + y^3 = cz^3. $$ You are asking for a general criterion for when $E_c(\mathbb Q)$ has positive rank. The Birch-Swinnerton-Dyer conjecture says this is true if and only if $L(E/\mathbb Q,1)=0$. Using the 3-isogeny coming from the cyclic subgroup generated by the points with $z=0$, maybe one can come up with a congruence condition on $c$ (at least when $c$ is prime?) that ensures the functional equation has odd sign, and then Birch-Swinnerton-Dyer would predict positive rank. And possibly one could use a Heegner point construction to actually produce a point of infinite order in some of those situations. In any case, given a 1-parameter family of elliptic curves over $\mathbb Q$ with generic rank $0$, the question of which parameter values give positive rank is generally quite subtle. So if you could pin down more precisely what sort of characteristics you're looking for, it would help.
Addendum: If you take $c$ so that the generic fiber has positive rank, you can get lots of examples with infinitely many solutions. For example, if you let $c=a^3+1$ with $a\in\mathbb Z$, then $[a,1,1]\in E_{a^3+1}(\mathbb Q)$, and $$ E_{a^3+1}(\mathbb Q)=\infty\quad\text{ for all but finitely many $a\in\mathbb Z$.} $$ (Probably this is true for all $a\notin\{-1,0,1\}$, but I haven't checked.) In any case, this gives lots of counterexamples to your guess, since very few values of $a^3+1$ will equal a prime power.
This is just a rough draft.
One example which has infinitely many coprime integer soutions using obvious soution.
$$A^3+B^3=cC^3\tag{1}$$
Let $x=A/C$, $y=B/C$ then we get equation $(2).$
$$x^3+y^3 = c\tag{2}$$
Cubic curve $(2)$ can be transformed to ellipric curve $(3).$
$$Y^2 = X^3-432c^2\tag{3}$$
$X = \frac{\large{12c}}{\large{x+y}}$
$Y = \frac{\large{36c(x-y)}}{\large{x+y}}$
$x = \frac{\large{36c+Y}}{\large{6X}}$
$y = \frac{\large{36c-Y}}{\large{6X}}$
We know a obvious soution $(x,y,c)=(m+1, -m, 3m^2+3m+1).$
Hence we get the rational point $P(X,Y)=( 36m^2+36m+12, 36(3m^2+3m+1)(2m+1)).$
This point is infinite order and $nP,n=2,3,...$ give infinitely many points.
The point $nP$ pulls back to the infinitely many integer solutions of equation $(1).$
For instance with $n=2$ and $3.$
$2P(X,Y)=\left( \frac{\large{12(1+7m^2+4m+3m^4+6m^3)}}{\large{(2m+1)^2}}, \frac{\large{36(-1-5m-9m^2+14m^4-2m^3+6m^6+18m^5)}}{\large{(2m+1)^3}}\right).$
$2P(X,Y)$ pulls back to $(A,B,C)=((m^3+6m^2+6m+2)m, -(m+1)(m^3-3m^2-3m-1), (m^2+m+1)(2m+1)).$
$3P(X,Y)=\left( \frac{\large{4(m^6+3m^5+12m^4+19m^3+15m^2+6m+1)}}{\large{(m+1)^2m^2}}, \frac{\large{4(2m^9+9m^8-18m^7-105m^6-207m^5-234m^4-165m^3-72m^2-18m-2)}}{\large{(m+1)^3m^3}}\right).$
Similarly, $(A,B,C)=( m^9+18m^8+45m^7+33m^6-36m^5-90m^4-78m^3-36m^2-9m-1, -m^9+9m^8+63m^7+138m^6+171m^5+144m^4+87m^3+36m^2+9m+1, 3m(m^6+3m^5+12m^4+19m^3+15m^2+6m+1)(m+1)).$
Thus, we can get the coprime solutions by removing the common factors.
Numerical example with $m=1$:
$c=3m^2+3m+1.$
$c = (7, 19, 37, 61, 91, 127, 169, 217, 271, 331,....)$
$c=7$
$A^3+B^3=7C^3$
$Y^2 = X^3-432c^2$
(A,B,C)
(2,-1,1)
(5,4,3)
(-17,73,38)
(-1256,1265,183)
(90271,-65882,40049)
(9226981,4381019,4989780)
(191114642,4309182809,2252725111)
(-2452184545855,2596383146704,733037580903)
(6782875656593327,-6048760527515143,2349209147442082)
(26028958492372169876,4925537406304613275,13637510581130984157)
