Number theory existential proofs Does there exist $2$ distinct primes $p$,$q$ such that 
$2^a$$p^b$=$q^c-1$ has no solution over natural numbers $a$,$b$,$c$.
 A: Depending on the details of the question, the answer is trivial but for opposite reasons. If you really mean "over the natural numbers", the answer is no - because $a = b = c = 0$ will be a solution no matter what $p,q,r$ are. On the other hand, if you mean "over the positive natural numbers", the answer is yes, all of them - no matter what $p,q,r$ are, $p^aq^b=r^c$ would give two different prime factorizations of the same integer, which is not possible.
Given that both these answers are sort of unsatisfying, I have to ask: are you sure you transcribed the problem you're interested in correctly?
EDIT: Your revised problem also has the answer yes, but now it's not entirely trivial. Note that $r^c - 1$ is always divisible by $r - 1$. So we just need to choose $p, q,$ and $r$ so that $r - 1$ is divisible by some prime that is not $p$ or $q$. Taking $p = 3$, $q = 5$, and $r = 17$ is one option that's particularly appealing to me. $17^c - 1$ is divisible by $16$ for every $c$, so if we had $3^a5^b = 17^c - 1$, we would have to also have that $16$ divides $3^a5^b$. But $3^a5^b$ is not even divisible by $2$, let alone by $16$.
EDIT 2: The latest change to the question really isn't significant, it just rules out the one option I chose in my previous answer. Again, $q^c - 1$ is divisible by $q - 1$, so we just need $q - 1$ to have a factor that isn't $2$ or $p$. Take $p = 3$ and $q = 11$. Suppose for contradiction that $2^a3^b = 11^c - 1$. Then $2^a3^b$ is divisible by $11 - 1 = 10$, so in particular must also be divisible by $5$. But $2^a3^b$ is clearly not divisible by $5$.
