To solve $2\cdot(5^y)-7^x=1$ How do we find all non-negative integers $x,y$ such that $2\cdot(5^y)-7^x=1$ ? 
Does there exist any solution with $x>2, y>2$ ? 
 A: For integers $a,b>0$ and $k\neq 0$, Pillai's conjecture states that the exponential Diophantine equation
$$
ax^m-by^n=k
$$
has only finitely many integer solutions $x,y,m,n>1$ with $nm\ge 6$. For many specific cases the conjecture has been proved, and even all solutions have been determined. For our case, $a=2$, $b=1$,$x=5$, $y=7$,$k=1$, we have $2\cdot 5^m-7^n=1$. It is shown in the article of Scott and Styer, that there are at most two positive integer solutions, see Corollary to Theorem $1$.
A: If $y \geq 3$, then taking $\pmod{125}$, $7^x \equiv -1 \pmod{125}$ so $x \equiv 10 \pmod{20}$. Write $x=10a$ where $a$ is odd. Then $2(5^y)=7^x+1 \equiv 7^{10a}+1 \equiv (-1)^a+1 \equiv 0 \pmod{7^{10}+1}$. However $281 \mid 7^{10}+1$, so we get a contradiction.
Thus there are no solutions with $y \geq 3$. We may check $y=0, 1, 2$ directly.
A: $\bmod 4$, then $2-(-1)^x\equiv 1\pmod4$, $x$ is even;
$\bmod 3$, then $2(-1)^y-1\equiv 1\pmod3$, $y$ is even.
so 
$$(7^x)^2-2(5^y)^2=-1$$
Consider Pell equation 
$$a^2-2b^2=-1$$
the fundamental solution is $(1,1)$, so 
$$a_n+b_n\sqrt2=\Big(1+\sqrt2\Big)^{2n+1}$$
next, find $s,t$ such that $a_n=sa_{n-1}+ta_{n-2}$,...
