Exponential Diophantine

To start, notice that it must be that $x,y \geq 0$ because if $x$ is negative then $2^x < 1 < 3^y + 5$ . And if $y$ is negative $3^y + 5$ won't be an integer so it can't equal $2^x$ . But also $2^0 < 3^0 + 5 \leq 3^y + 5$ so $x,y > 0$ . This means only natural solutions can exist.

I have no clean way of finding solutions, but I found a way to reduce the number of possibilities I would have to check. First, arrange the equation like $3^y - 2^x = -5$ and notice this implies the weaker equation $|3^y - 2^x| = 5$ . This gives us the following idea. If we can show that $|3^y - 2^x|$ grows larger than 5 then that means we would only have to check a finite amount of cases. So let's start bounding:

For a given $y$ , let $x_y$ be the biggest number such that $2^{x_y} < 3^y$ . This implies that $2^{x_y + 1} > 3^y$ and then this implies the inequality: $|3^y - 2^x| \geq \min \{ |3^y - 2^{x_y}|, |3^y - 2^{x_y + 1}| \}$ . Let's now study the properties of $x_y$ more closely. As I mentioned, we have $2^{x_y} < 3^y < 2^{x_y + 1} \implies x _y< y \log_2 3 < x_y+1 \implies 0 < y \log_2 3 - x_y < 1$ . For natural $x,y$ this implies that $x_y = \lfloor y \log_2 3 \rfloor$ . Replacing this into our inequality we have that:

$|3^y - 2^x| \geq \min \{ |3^y - 2^{\lfloor y \log_2 3 \rfloor}|, |3^y - 2^{ \lfloor y \log_2 3 \rfloor + 1}| \}$ . It can be easily proven that both of the functions inside the $\min$ are increasing functions and for $y > 3$ we already have $\min \{ |3^y - 2^{\lfloor y \log_2 3 \rfloor}|, |3^y - 2^{ \lfloor y \log_2 3 \rfloor + 1}| \} > 5$ . So we only need to consider $y \leq 3$ and because of the first deductions of this solution we know we just have to check $y=1,2,3$ .

Doing this, we find two solutions. $2^3 = 3^1 + 5$ and $2^5 = 3^3 + 5$ .