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.
 A: 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.
A: With the two substitutions:
$$x-4=w$$
$$y=2z$$
We rearrange to:
$$5(2^w)=(3^z-1)(3^z+1)$$
So we want $z$ such that $$\bigg[3^z-1=5(2^a)\bigg] \text{ and } \bigg[3^z+1=2^b\bigg]$$ 
or
$$\bigg[3^z+1=5(2^a)\bigg] \text{ and } \bigg[3^z-1=2^b\bigg]$$ 
We can see $z=2$ (and thus $y=4$) satisfies the second line here. (which leads to $x=8$ as others conclude).
See if you can take it from here.
A: From $3^y\equiv1$ mod $5$, we see that $4\mid y$, so writing $y=4z$, we have
$$5\cdot2^{x-4}=3^{4z}-1=(3^{2z}+1)(3^{2z}-1)$$
Since $3^{2z}+1\equiv2$ mod $8$, we can only have $3^{2z}+1=2$ or $10$.  We can dismiss the first possibility (since it implies $z=0$, which gives $3^{2z}-1=0$). Thus $3^{2z}+1=10$, so $z=1$ and thus $5\cdot2^{x-4}=10\cdot8$ implies $x-4=4$. We find $(x,y)=(8,4)$ as the only solution.
A: Try $x=8$ and $y=4$. Not sure if there are any other integer solutions.
A: Solving for $y$ yields no solutions for $-3000\le x \le 3000$ but solving for $x$ does. WolframAlpha has a solution of $x$ in terms of $y$ here and plugging this into a spreadsheet does yield one solution in this range.
$$5\times 2^{x-4}-3^y+1=0\implies 
x = \frac{\log(\frac{16(3^y - 1)}{5})}{\log(2)}\implies x=8\land y=4$$
