The diophantine equation $5\times 2^{x-4}=3^y-1$ I have this question: can we deduce directly using the Catalan conjecture that the equation
 $$5\times 2^{x-4}-3^y=-1$$
has or no  solutions, or I must look for a method to solve it. Thank you.

Elementary proof. I learned the method at Exponential Diophantine equation $7^y + 2 = 3^x$ 
We think that the largest answer is $5 \cdot 16 = 81 - 1. $ Write this as 
$5 \cdot 16 \cdot 2^x = 81 \cdot 3^y - 1.$ Subtract $80$ from both sides, 
$ 80 \cdot 2^x - 80 = 81 \cdot 3^y - 81.$ We reach
$$ 80 (2^x - 1) = 81 (3^y - 1).  $$
This is convenient; we will show that both $x,y$ must be zero. That is, ASSUME both $x,y \geq 1.$
From $2^x \equiv 1 \pmod {81}$ we get
$$ x \equiv 0 \pmod {54}. $$
It follows that $2^x - 1$ is divisible by $2^{54} - 1.$
$$ 2^{54 } - 1 = 3^4 \cdot 7 \cdot 19 \cdot 73 \cdot 87211 \cdot  262657
$$
Next, $3^y - 1$ is divisible by the large prime $262657$
From $3^y \equiv 1 \pmod {262657}$ we find
$$ y \equiv 0 \pmod {14592} $$ and especially
$$ y \equiv 0 \pmod {2^8}. $$
We do not need as much as $2^8 = 256,$ we really just need the corollary
$$ y \equiv 0 \pmod 8  $$
Next $3^y - 1$ is divisible by $3^8 - 1 = 32 \cdot 5 \cdot 41.$  This is the big finish, $3^y - 1$ is divisible by $32.$ Therefore $80 (2^x-1)$ is divisible by $32,$ so that $2^x - 1$ is even. This is impossible if $x \geq 1,$ and is the contradiction needed to say that, in
$$ 80 (2^x - 1) = 81 (3^y - 1) \; ,  $$
actually $x,y$ are both zero.
