2023/004) Solve in integers $4^x-5^y= 39$

we have $4^x = 2^{2x}$

now working in mod 3 we get $1-5^y \equiv 0 \pmod 3$

or $5^y \equiv 1 \pmod 3$

as we know $5 \equiv -1 \pmod 3$ so y has to be even say 2m

now $4^x - 5^y = 2^{2x} - 5^{2m} = 39$

Farctoring  we get $(2^x + 5^m)(2^x-5^m) = 39 = 39 *1 = 13 * 3$ (39 can be factored in 2 ways)

so we have 2 cases

$2^x+5^m= 39$ and $2^x - 5^m=1$ adding we get $2^x *2 = 40$ and this does not have integer

or 

$2^x+5^m= 13$ and $2^x - 5^m=3$ adding we get $2^x *2 = 16$ or x = 3 and subtracting $2 * 5^m = 10$ and m = 1

so x = 3 and y = 2