We start by changing the letters, plugging in your maximal exponents, and arriving at
$$ 128 (2^x - 1) = 125 (5^y-1) \; . \; $$
We ASSUME that both $x,y \geq 1$ and get a contradiction.
The main tasks are factoring $p^n - 1$ and finding the order of a prime for some target $n,$ the minimal $k$ such that $p^k \equiv 1 \pmod k$
So, order_2 of 125 is 100, where we have simply $\varphi(125) = 100.$
Thus $100|x.$
Next
$$ 2^{100} - 1 = 3 \cdot 5^3 \cdot 11 \cdot 31 \cdot 41 \cdot 101 \cdot 251 \cdot 601 \cdot 1801 \cdot 4051 \cdot 8101 \cdot 268501 $$
We need $5^y \equiv 1 \pmod {268501}$
Therefore $125 | y.$
$$ 5^{125} - 1 = 2^2 \cdot 11 \cdot 71 \cdot 101 \cdot 251 \cdot 401 \cdot 3597751 \cdot 9384251 \cdot \mbox{BIG} $$
We need $2^x \equiv 1 \pmod { 9384251}$
Therefore $125 | x.$ From the beginning, we have had $100 | x.$ Put them together, Chinese Remainder Theorem, and we arrive at
$$ 500 | x $$
And
$$ 2^{500}-1 = (2^{100}-1)(2^{400}+2^{300}+2^{200}+2^{100}+1)$$
where $2^{100}-1$ is a multiple of $5^3$ (above) and the second factor is a multiple of $5$ because $2^{100}=16^{25}\equiv 1\bmod 5$. Thus $2^{500} - 1$ is a multiple of $5^4.$ We have gotten what we wanted, as
$ 128 (2^x - 1) = 125 (5^y-1) \; \; $ tells us that $ 125 (5^y-1)$ is divisible by $625.$ In turn, this says that $ (5^y-1)$ is divisible by $5.$ This is impossible with $y \geq 1,$ so that $y = 0,$ contradicting our hypothesis.
factor with gp-pari
parisize = 4000000, primelimit = 500000
? factor( 2^100 -1)
%1 =
[ 3 1]
[ 5 3]
[ 11 1]
[ 31 1]
[ 41 1]
[ 101 1]
[ 251 1]
[ 601 1]
[ 1801 1]
[ 4051 1]
[ 8101 1]
[268501 1]
? factor( 5^125 -1)
%2 =
[ 2 2]
[ 11 1]
[ 71 1]
[ 101 1]
[ 251 1]
[ 401 1]
[ 3597751 1]
[ 9384251 1]
[ 28707251 1]
[ 4032808198751 1]
[ 767186663625251 1]
[24687045214139234043375683501 1]
? factor( 2^500 -1)
%3 =
[ 3 1]
[ 5 4]
[ 11 1]
[ 31 1]
[ 41 1]
[ 101 1]
[ 251 1]
[ 601 1]
[ 1801 1]
[ 4051 1]
[ 7001 1]
[ 8101 1]
[ 28001 1]
[ 96001 1]
[ 268501 1]
[ 3775501 1]
[ 229668251 1]
[ 269089806001 1]
[ 4710883168879506001 1]
[ 47970133603445383501 1][+++]
?
