Solutions of $10^{x} -5^{x}-2^{x}=y^{2}-1$. How do i show that the only positive integral solutions to the equation $$10^{x} -5^{x}-2^{x}=y^{2}-1$$ are $(x,y)=(1,2)$.
Tried a lot of things. This looks quite difficult and i am not even sure how to start.

Here I discus a certain case.
$10^x-2^x-5^x=y^2-1$
If $p= x+1$ is prime  we may write:
$10^{p-1} ≡ 1 \mod p$
$2^{p-1} ≡1 \mod p$
$5^{p-1} ≡1 \mod p$
$10^{p-1}-2^{p-1}-5^{p-1}≡ -1 \mod p≡ (p-1)\mod p$
$10^{p-1}-2^{p-1}-5^{p-1}= t . p -1= y^2-1 ⇒ y=t.p$
This is possible only if $t=p$ ⇒ $y=±p$. For example $x=1⇒t= p=2 ⇒y=±2$
in this case question reduces to:
$10^x-2^x-5^x=(x+1)^2-1$
or
$10^{p-1}-2^{p-1}-5^{p-1}=p^2-1$
which are 'single unknown' equations.
We may also write:
$10^{p-1}-2^{p-1}-5^{p-1}= t . p + p -1= y^2-1 ⇒ y=t(p+1)=t(x+2)$
Since (p+1) is not prime if t completes factors of (p+1) for a square
then we have an integer solution for y. It is not known in addition to 2 there are other primes that would give integer solutions for y.
