The diophantine equation $5\times 2^{x-4}=3^y-1$

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.

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.

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.