Integer solutions to $2^m - 3^n = p \cdot C$ where $m, n , p$ are positive integer variables and $C$ is a positive integer constant. How many solutions are there for equation  $2^m  -  3^n = p \cdot C$ where $m, n, p$ are positive integer variables and $C$ is an odd positive integer constant greater than $3$?
Can we say that if there will be none, one, finitely many, or infinite number of solutions?
What happens if $C$ is prime? We believe that for prime $C$, there is at least one solution. Can we prove it?
 A: Bennett has proved the following result, see here:
Theorem: Let $a,b,c$ be positive integers with $a,b\ge 2$. Then
$$
a^m-b^n=c
$$
has at most two positive integer solutions $(m,n)$.
For example, the equation $2^m-3^n=5$ has the two solutions $(m,n)=(3,1),(5,3)$.
A: IF $p$ is variable integer, then for a pair $(m,n)$ there exists some $p$ such that
$$2^m-3^n=pC$$
if and only if
$$2^m=3^n \pmod{C}$$
It is also easy to argue that there are infinitely many solutions of this second equation.
If $p$ is fixed, the other answer addressed this.
A: There are infinitely many solutions. Let $C=11$, $n=1$ and $m=10l+8,\exists   l\in\mathbb{N}$ then $$2^{m}-3^n\equiv 2^{10l+8}-3\equiv 2^8-3\equiv 0\pmod{11}$$
and thus, we can let natural number $p=\dfrac{2^{10l+8}-3}{11}$ so that $(m,n,p,C)=(10l+8,1,1,11)$ are the solutions to the equation $2^m-3^n=11k$ $\Box$

And when $C$ is prime we can let $m= (C-1)q_1$ , $  n=(C-1)q_2$ for some natural number $q_1,q_2$ so that, by Fermat's little theorem
$$2^m-3^n\equiv1-1\equiv 0\pmod {C}$$
and choose $p=\dfrac{2^{(C-1)q_1}-3^{(C-1)q_2}}{C}$ that are the solutions to $2^m-3^n=pC$ where $C$ is prime, Hence, again there is infinitely many solution for prime $C$.
