Prove there is no rational number r such that $2^r = 3$

There are several logical mistakes in your argument:

($1$) $2^p=3^q$ does not imply $3^q$ is a multiple of $2$. You need to consider the case $p=0$ and $p \neq0$ separately.

($2$) "In the case that $2^p$ is odd we have an even number being equal to an odd number which is also a contradiction"

This statement is not correct. When $p=0$, we have an odd number equals an odd number. The fallacy comes from the assumption in ($1$)

Edit: as comment by Graham Kemp points out, you also need to prove the case where $p<0$ and $q>0$, which is trivial but need to be stated.

You need to consider if $r < 0$. And you need to redo if $r = 0$ correctly.

If $r > 0$ then there are $p, q \in \mathbb Z^+$ where $r =\frac pq$ and, although we can claim $p, q$ have no common factors that is not relevant or necessary. Your argument was PERFECT. $2^{\frac pq} =3 \implies 2^p = 3^q$ but LHS is even and RHS is odd. Beautiful!

If $r = 0$ you kind of botch it. You say $2^0 = 3^q$ so $3^q = 1$ is odd which.... is not a contradiction. More to the point: If $r = 0$ then $2^r = 2^0=1$ which.... is not equal to $3$ That's all there is to it.

And to consider $r < 0$, if $r < 0$ then there are $p,q \in \mathbb Z^+$ so that $r =-\frac pq$ so $2^{-\frac pq} = \frac 1{2^{\frac pq}} = 3$ so raise both sides to the $q$ power and get $\frac 1{2^p} = 3^q$ and LHS is less than $1$ while RHS is more than $1$.