$c^2(a\cdot b)+c(a+b)=2^c-2$ is of the form $mx+ny=k$ and should open the door to Diophantine, but do the constraints $x=(a\cdot b), y=(a+b)$ make a difference when trying to solve for $\mathbb{N}$ solutions?
For example:
$11^2(2\cdot8)+11(2+8)=2^{11}-2=2046$
However, $c=9$ does not seem to have any solutions where $a,b \in \mathbb{N}$.
For both 9 and 11 I just brute forced an answer, but if I wanted to try and find positive, non-zero integer solutions to a large number c, is there a more systematic way of going about it?
Note that $c^2ab + ca+cb = (ca+1)(cb+1)-1$
Thus we are trying to solve
$$\textbf{I)} \text{ }\text{ }\text{ }\text{ }(ca+1)(cb+1)=2^{c}-1.$$
This equation has solutions for a given $c \in \mathbb{N}$ iff $2^{c}-1$ is a product of two factors $u,v$ such that both are $1 \mod c$. A necessary condition is $2^{c}-1 \equiv 1 \mod c$. If your definition of $\mathbb{N}$ includes $0$ then the previously mentioned condition is also sufficient.
For your special case $c = 9$ we have $2^{9}-1 = 511$ which is $7 \mod 9$ and thus $\textbf{not}$ a product of two numbers which are $1 \mod 9$, hence there will be no solutions.