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?

As you noted: $y =\frac {2^{2k} - 1}3$ is always an integer.  So $2y = 2\frac {2^{2k} - 1}3 = 2\frac {2^{2k+1} - 2}{3} = \frac {2^{2k+1} - 1}3 - \frac 13$ is integer so $\frac {2^{2k+1} -1}3$ is not an integer.
So no.
If know modulo arithmetic it's easier.
$y = \frac {2^{2k} -1}3$ being an integer is the same as $3y +1 = 2^{2k}$ is the same as saying $2^{2k}$ has remainder $1$ when divided by $3$ which we write as $2^{2k} \equiv 1 \mod 3$.  ($a \equiv b \mod n$ means $a$ and $b$ have the same remainder when divided by $n$).
So $2*2^{2k} = 2^{2k + 1}$ will have remainder $2*1 = 1$ when divided by $3$ or $2^{2k+1} = 2*2^{2k} \equiv 2*1 = 2 \mod 3$ and $2^{2k+1} \not \equiv 1 \mod 3$  So $\frac {2^{2k+1} - 2}3$ is an integer but $\frac {2^{2k+1} - 1}3$ is not.
In general though $\frac {2^{2k} -1 }3$ is an integer; $\frac {2^{2k+1} + 1}3 $ is an integer. And we can generalize that as $\frac {2^{x} + (-1)^x}3$ is an integer.
Or in modulo notation:
$(2)^x \equiv (-1)^2 \mod 3$ which makes sense as $2 \equiv -1 \mod 3$ [$2 = 0*3 + 2$ with $2$ remainder and $-1 = -1*3 + 2$ with $2$ remainder.]  So $2^x = (3-1)^x = 3^x - x*3^{x-1}......+ x*3*(-1)^{x-1} + (-1)^x$ will have the same remainder as $(-1)^x$.
