Answer: $(a,b)=(2,2)$ Notice that for $7^a\equiv 5^b$ we must have both $a$ and $b$ being even. Let $a=2a'$ and $b=2b'$ then we must have that $$\frac{49^{a'}-25^{b'}}{8}=p$$But as $3$ divides the top we must have that $p=3$ or that $(7^{a'}+5^{b'})(7^{a'}-5^{b'})=24$. But $7^{a'}-5^{b'}$ is at least $2$ and $7^{a'}+5^{b'}$ is at least $12$ so both must be equalities leading to the above solution.

$49^x-25^y=24$ Since $A(x,y)=49^x-25^y=(7^x-5^y)(7^x+5^y)=24$ checking the cases gives $(a,b)=(2,2)$