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.
 A: According to Catalan's conjecture (or Mihăilescu's theorem), there are no non-trivial solutions.
The theorem states that the only solution in the natural numbers of
$x^a − y^b = 1$ for $a, b > 1, x, y > 0$ is $x = 3, a = 2, y = 2, b = 3.$
link for the wiki page.
Despite of this powerful theorem, there is an easier way to solve this specific equation.
If $y=1$, then $x=2$. If $y >1$, then $6|x$ by taking modulo $9$. This will imply $2^x-1$ is divisible by $7$.
A: This particular equation is a special case of Mihailescu's theorem: rewriting it as $2^x - 3^y = 1$, we know that either $x \le 1$ or $y \le 1$, because the only instance of higher powers having difference $1$ is $3^2 - 2^3$. This gives us the solutions $(x,y) = (2,1)$ and $(x,y) = (1,0)$.
In general, exponential Diophantine equations like this one can be solved by tedious application of the technique "take the equation modulo some arbitrary number like $37$ or something".
It's not quite that bad, though: we would just look for values of $m$ such that $2$ or $3$ have small multiplicative order mod $m$, i.e., $m$ divides $2^k-1$ or $3^k-1$ for some small $k$. This would either tell us that no solutions exist or give us some modular conditions on $x$ and $y$ which form part of a contradiction. 
If some small solutions do exist, we would have to consider the equation modulo $2^k$ or $3^k$ to eliminate larger solutions. For example, in this problem, we have solutions for $x=1$ and $x=2$. If we had a solution in which $x \ge 3$, we would have $2^x \equiv 0 \pmod 8$. In that case, $$3^y \equiv 2^x - 1 \equiv 7 \pmod 8$$ but this is impossible: $3^y \bmod 8$ is $3$ when $y$ is odd, and $1$ when $y$ is even. So there can be no solution with $x \ge 3$.
(It's easy to check $x=0$, $x=1$, and $x=2$ to figure out that the above solutions are the only ones possible.)
A: 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 $(x,y)=(2,1)$.
