Solve over the positive integers: $7^x+18=19^y.$ Solve over the positive integers:
$$7^x+18=19^y.$$
Progress:-
I first took $\mod 7,$ so we get $5^y\equiv 4 \mod 7$ since $5$ is a primitive root of $7$ and $5^2\equiv 4\mod 7.$ So we get $y\equiv 2\mod 6.$
And then took $\mod 9$
So we get $7^x\equiv 1\mod 9.$ Since residues of $7^x$ are $\{7,9,1\}.$
We get $x\equiv 0\mod 3.$
Then I couldn't get any progress, I tried Zsigmondy,etc stuff and also noticed $7^x-1=19(19^{y-1}-1)$
Any hints? Thanks in advance.

Excluding the obvious solution $(x, y) = (0, 1)$, an argument mod $7$ shows that $y$ is even (which you have done). Similarly, an argument mod $19$ shows that $x$ is a multiple of $3$.
Thus we may look at the elliptic curve $Y^2 = X^3 + 18$. A computer algebra system such as Sage can be used to find all integral points: $(X, Y) = (7, \pm 19)$.
Therefore the only remaining solution is $(x, y) = (3, 2)$.

There might be elementary solutions, but this is the most efficient way of solving it.
