<s>[INST] Will there always be integer solutions for $x$ and $y$ in $3^y = 2^x - 1$? Is there a way to find the solutions of $3^y=2^x-1$ where $(x,y)$ are integer coordinates (maybe within a certain interval)? Are there an infinite number of integer coordinate pairs for this equation? Is it possible to even determine this? I'm not quite sure the difficulty of this problem, but substituting $x$ or $y$ values into this equation to find integer pairs is time consuming and not very promising.
 [/INST] There is an entirely elementary way to show that $(x,y)=(1,0)$ and $(2,1)$ are the only integer solutions to $3^y=2^x-1$.
If $y\ge1$, $x$ must be even since $2^{2u+1}-1\equiv1$ mod $3$.  Writing $x=2u$, we have
$$3^y=2^{2u}-1=(2^u-1)(2^u+1)$$
which implies $2^u-1$ and $2^u+1$ are each a power of $3$, say $2^u-1=3^r$ and $2^u+1=3^{r+s}$ with $s\ge1$ (since $2^u+1$ is clearly greater than $2^u-1$).  This gives us 
$$2=(2^u+1)-(2^u-1)=3^r(3^s-1)$$
the only solution to which is $r=0$ and $s=1$, leading to $u=1$, which is to say $(x,y)=(2,1)$.
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