This is a proof of a special case of [Goormaghtigh's conjecture][1],namely
$$\frac{10^m-1}{10-1} = \frac{y^n-1}{y-1}$$ satisfying $10>y>1$ and $ n,m >2 $ has no non-trivial solution in non-negative integers.
Proof:
$$\frac{10^{m-1}-1}{10-1}.\frac{10}{y} = \frac{y^{n-1}-1}{y-1}$$
$\therefore10 |y$ or $(10^{m-1}-1)|y\ \because \frac{y^{n-1}-1}{y-1}$ is in naturals set.
$$\implies y = 2,3,5,9,I_{m-2}$$
By observation, $ y \neq I_{m-2}$
Case 1 $ y = 9$
$$\implies\frac{10^m-1}{10-1} = \frac{9^n-1}{8} $$
$$\implies \ 8*10^m = 9^{n+1}-1$$
It is easy to see that this has no solution for the given conditions
$\therefore \ y \neq 3$
Case 2 $ y = 5$
$$\implies\frac{10^m-1}{10-1} = \frac{5^n-1}{4} $$
$$\implies \ 4*10^m = 9*5^n-5$$
$$\implies \ 8*10^{m-1} = 9*5^{n-1}-1$$
It is easy to see that this has no solution for the given conditions
Case 3 $ y = 2 $
$$\implies\frac{10^m-1}{10-1} = 2^n-1 $$
Similarly for case 3
Hence proved
I understand that there is not much detail regarding the 'it is easy to show', but I have merely posted the proof sketch to know if my strategies work.
I am a high schooler and am not fluent in mathematical proof-writing. I understand that there could be flaws or misunderstanding and will be more than glad to receive opinion/feedback/advice on this proof attempt
PS $I_{m-2}$ is the repunit of order $m-2$ where the repunit of order $3$ is $111$ and so on
Looking at this I see that I made a fatal flaw not considering prime factors of $I_{m-2}$