There is an arithmetic sequence $2^a, 3^b, 4^c$ such that a,b and c are positive integers. The question posed is to find ALL possible ordered triplets $(a,b,c)$.
The constant difference d between the terms allows for a system of 3 equations with 4 unknowns, but properties of the odd and even bases allows deduction of a in the following way:
$3^b - 2^a = d$
$4^c - 3^b = d$
Eliminating d and rearranging gives
$2^{2c-1} + 2^{a-1} = 3^b$
Which can only be true in the cases where a=1 because of the integer clause. So now what is left is the relationship
$2^{2c-1} + 1 = 3^b$
Which implies that this works for any power of 2 and power of 3 which differ by 1. I can think of no way to show how to find the pairs 2,3 or 8,9 through any reasoning other than intuition, and thus I don't know how to say that there wouldn't be some unknowingly large pair that would also work.
As you say, our problem is equivalent to: Solve in positive integers: $$2^{2c-1}+1=3^b$$
$(c,b)=(1,1)$ is a solution. Let $c\ge 2$.
Then mod $4$ gives $b=2m$ for some $m\in\Bbb Z^+$.
$$2^{2c-1}=\left(3^m+1\right)\left(3^m-1\right)$$
LHS is a power of $2$, so $3^m+1$ and $3^m-1$ are both too.
But $\left(3^m+1\right)-\left(3^m-1\right)=2$, so $m=1$.
Answer: $(c,b)=(1,1),(2,2)$.