# Elementary solution to $4^x-25^y=39$ in natural numbers

How many natural pairs $(x,y)\in\Bbb{N^2}$ exist such that $4^x-25^y=39$ holds?

The answer is $1$ the trivial $(x,y)=(3,1)$ is the only one; I've found Catalan's conjecture generalizations for $n=39$ but there should be an elementary solution.

This was given as one of $20$ questions from entrance exam for university of civil engineering which usually have pretty elementary solutions.

If anyone is curious this is how the exam looked like: Exam (it's not in English though)

Since $(2^x-5^y)(2^x+5^y)=39$, $(2^x-5^y,2^x+5^y)=(1,39)$ or $(3,13)$.
So, $(2^x,5^y)=(20,19)$ or $(8,5)$.
It is impossible that $(2^x,5^y)=(20,19)$.
$(x,y)=(3,1)$.