Find integers $x$ and $y$ such that $8^x-9^y=431$ Find integers $x$ and $y$ such that $8^x-9^y=431$
My working:
By taking mod 9 and 16, I got $x$ odd and $y$ even.
Also $8^x>431\implies x\ge 3$
For $x=3$ I got $y=2$

SECOND ANSWER: We can stick with smaller primes in the other direction
We suspect that $512-81$ is the largest solution. Proof by contradiction:
Giving new names to $x,y,$  we say
$$  512(8^x - 1) = 81 (9^y - 1)  $$
We ASSUME both $x \geq 1, y \geq 1.$
First, $9^y \equiv 1 \pmod {512}.$ A calculation shows that $y$ must be divisible by $64$
Next,
$$ 9^{64} - 1 = 3^{128} - 1 = 2^9 \cdot  5 \cdot 17 \cdot 41 \cdot 193 \cdot ... $$
wWe use the 193.
$$ 8^x \equiv 1 \pmod{193} $$
so that $x$ is divisible by $32$ 
Then $8^x - 1$ is divisible by 
$$8^{32} - 1 = 2^{96}-1 = 3^2  \cdot 5  \cdot 7 \cdot 13 \cdot 17 \cdot 97 \cdot 193 \cdot 241 \cdot 257 \cdot 673 \cdot 65537 \cdot  22253377
$$ 
We notice some Fermat primes here, in particular $257 = 2^8 + 1$ is a factor of $2^{16} - 1,$ in turn this divides $2^{96} - 1.$
That's all we needed. We find $9^y-1$ divisible by $257.$ 
This tells us that $y$ is divisible by $128.$ But $9^{128} - 1 = 3^{256}-1$ is divisible by 1024.
Therefore $512(8^x - 1) $ is divisible by $1024,$ which is a contradiction of $x \geq 1.$
