Solve $2^x+3^y=z^2$ in nonnegative integers. So, we are trying to find all the solutions to $2^x + 3^y = z^2$ in nonnegative integers. Here are my insights: 
First of all, $z^2$ can be either $0$ or $1$ modulo $3$. If $z^2 = 3k$, then
LHS cannot be divisible by $3$ unless $y = 0$ and $x=2k+1$, and so we have $2^x+1 = z^2$, so one of the solutions is $(3,0,3)$ (but we still need to show that this is the only solution in this case). 
Now let $y > 0$, $x=2k$. In this case $2^x \equiv 1\ (\mod\ 3)$ and $z^2 \equiv 2^x (\mod 3)$. We have
$$
3^y = (z-2^k)(z+2^k),
$$
so $(z-2^k)=3^a$ and $(z+2^k)=3^b$, so $3^b-3^a = 2^{k+1}$. If $a > 0$, then
$2^{k+1}$ must be divisible by $3$, but it is not, so suppose $a = 0$. 
Now we have
$$
2^{k+1} = 3^b - 1,
$$
Аnd this is where I stuck. It is easy to find two more solutions just iterating over $k$, but I do not know how to prove that there are no others. As far as I understand, we need to find some kind of an upper bound for $k$. Note: I am trying to avoid the use of Catalan's conjecture here.
Can you please help me to show that there are no other solutions to this equation and that $2^x+1=z^2$ has a unique solution?

The second question answer:
$2^x = z^2 - 1 \Rightarrow 2^x = (z+1)(z-1)$. If $x$ and $z \in \mathbb N$, then this can be true only for $z=3$ as difference between $z+1$ and $z-1$ equals $2$ and we have no other degrees of $2$ which are different by $2$.
