$7^x-3^y=4$ in the integers I have to find all solutions to $7^x-3^y=4$ in the integers, I've already proven that $x$ and $y$ have the same parity and that they cannot be even. But I'm stuck in the case when $x$ and $y$ are odd. Could someone show me how to solve for all solutions? I know there is at least one, $(x,y)=(1,1)$, but I don't know if there's any other solution

If $y\leq 1$, the only solution is $(1,1)$, which you found. If $y\geq 2$, then consider the equation $\bmod 9$. We see
$$7^x\equiv 4\bmod 9$$
We have
$$7^0\equiv 1,\ 7^1\equiv 7,\ 7^2\equiv 4,\ 7^3\equiv 1,$$
so $7^x\equiv 4\bmod 9$ iff $x\equiv 2\bmod 3$. As a result, we have $x\equiv 5\bmod 6$ (as you have already shown $x$ is odd). So, as
$$7^{12}\equiv 1\bmod 13,$$
we have
$$7^x\equiv 7^5\mathrm{\ or\ }7^{11}\equiv \pm 2\bmod 13$$
since $x\equiv 5\bmod 6\implies x\equiv 5\mathrm{\ or\ }11\bmod 12$. This means that
$$3^y=7^x-4\equiv (\pm 2)-4\in\{7,11\}\bmod 13.$$
However
$$3^0\equiv 1,3^1\equiv 3,3^2\equiv 9,3^3\equiv 1\bmod 13,$$
so we get a contradiction. 
