Integer solutions to $y=(2^x-1)/3$ where $x$ is odd For the equation $y=(2^x-1)/3$ there will be integer solutions for every even $x$.
Proof: When $x$ is even the equation can be written as $y=(4^z-1)/3$ where $z=x/2$.
$$4^z =1 + (4-1)\sum_{k=0}^{z-1} 4^k$$
If you expand that out you get:
$$4^z=1+(4-1)4^0+(4-1)4^1+(4-1)4^2+\dots+(4-1)4^{z-2}+(4-1)4^{z-1}$$
Which becomes:
$$4^z=1+4^1-4^0+4^2-4^1+4^3-4^2+\dots+4^{z-1}-4^{z-2}+4^z-4^{z-1}$$
After canceling everything out you are left with:
$$4^z=4^z$$
More generally:
$$a^z =1 + (a-1)\sum_{k=0}^{z-1} a^k$$
Therefore: $(2^x-1)/3$ will always be an integer when $x$ is even.
My question is: will there ever be an integer solution to $(2^x-1)/3$ when $x$ is odd?

If x is odd we can write x= k+ 1 for k an even integer.  Then $\frac{2^x- 1}{3}= \frac{2^{k+1}- 1}{3}= \frac{2(2^k)- 1}{3}= \frac{2(2^k)- 2+ 1}{3}= \frac{2(2^k- 1)+ 1}{3}= 2\frac{2^k- 1}{3}+ \frac{1}{3}$.
You have already shown that $\frac{2^k-1}{3}$ is an integer so this is an integer plus 1/3, not an integer.
