Different substitutions, let
$$ u = 5^x \; , \; \; v = 2^{y-1} $$
Then
$$ \frac{1}{4} \left( (2u+v-2)^2 - 49 v^2 \right) = 0 \; , \; \; $$
$$ \left( 2u+8v-2 \right) \left( 2u-6v-2 \right) = 0 \; , \; \; $$
$$ \left( u+4v-1 \right) \left( u-3v-1 \right) = 0 \; , \; \; $$
Both $u,v > 0$ so we are left with
$$ 5^x = 3 \cdot 2^{y-1} + 1 $$
For example, $3 \cdot 8 + 1 = 25.$
Monday: in some cases there is an elementary proof that we have already found the largest solution. We already have $3 \cdot 8 + 1 = 25.$ If we were to have a larger solution, it would be of the form
$$ 24 \cdot 2^s + 1 = 25 \cdot 5^t $$
Subtract 25 from both sides,
$$ 24 \cdot 2^s - 24 = 25 \cdot 5^t - 25 \; , \; $$
$$ 24 \left( 2^s - 1 \right) = 25 \left( 5^t - 1 \right) $$
We will ASSUME $s,t \geq 1$ and get a contradiction.
Since $25 | (2^s - 1),$ we find $2^s \equiv 1 \pmod {25}$ and then $s$ is divisible by 20. This
$$ 2^{20} - 1 = 3 \cdot 5^2 \cdot 11 \cdot 31 \cdot 41. $$
The actual $2^s - 1$ is divisible by $2^{20} - 1,$ and therefore by the prime $41.$
Since $41 | (5^t - 1),$ we find $5^t \equiv 1 \pmod {41}$ and then $t$ is divisible by 20. This
$$ 5^{20} - 1 = 2^4 \cdot 3 \cdot 11 \cdot 13 \cdot 41 \cdot 71\cdot 521 \cdot 9161. $$
The actual $5^t - 1$ is divisible by $5^{20} - 1,$ and therefore by $2^4 = 16.$
We have arrived at $$ 16 | 24 (2^s - 1). $$ Divide throsgh by $8,$ we get
$$ 2 | 3 (2^s - 1) $$
This is a contradiction, as both $3$ and $2^s - 1$ are odd when $s \geq 1.$ The contradiction tells us the assumptions are wrong, and
$$ s = 0 $$
This completes the proof that $3 \cdot 8 + 1 = 25$ is the largest solution.