Quite new here and after researching I haven't seen the thing I was looking for. Also, my apologies if I use wrong math variables, I am an enthusiast, not a professional.
The question is: For x= 1 to infinity and y = 1 to infinity, "in $2^x - 3^y = z$ , is $z$ increasing for any combination $x$ and $y$, while $z$ is positive ?" To further clarify below some examples.
Examples:
$2^5 - 3^3 = 5 $
$2^4 - 3^2 = 7$
$2^8 - 3^5 = 13$
...
$2^{152} - 3^{95} = 3.588096*10^{45}$
$2^{152} - 3^{96} = -6.536947*10^{44}$ negative and therefore not valid
$2^{153} - 3^{2} = 1.141798*10^{46}$
This continues at least for any combination for $x=450, y=283$. But is this for any value of $x$ and $y$? Or is there somewhere a difference that is stopping this trend and has a very tiny (positive) difference? For example: $2^{12345678} - 3^{123456} = 9.$
In other words, is there a proof that states that this trend is continueing?
A lot of thanks in advance.
EDIT:
There is a trend that for any $x$ combined with any $y$ the equation:
$2^x - 3^y = z$
where $z > 0$, $z$ will increase and never jump back to a (respectively) small difference.
Additional info since it is being flagged as unclear by some: Is there a $2^x$ slightly higher than $3^y$? (e.g. $100>z>1$ for a value $x > 8$: the lowest difference is $2^8-3^4=175$. OR $1000>z>1$ for a value $x > 11$: the lowest difference = $2^{11}-3^6=1319$) If there is a value $z$ for such a value $x$, that would make a big jump down.
@minus has prove the value $z$ will increase, see comment in answer below.
From what I gather, assuming we are working in the real numbers;
This is not true for all x and y.
$2^{x} - 3^{y} > 0$
$ xln(2) > yln(3)$
$x/y > ln(3)/ln(2)$
As shown by Minus one-twelth’s comment, you can also show that the further away from the ratio we get, the bigger the difference.