Are there more transcendental numbers or irrational numbers that are not transcendental? This is not a question of counting (obviously), but more of a question of bigger vs. smaller infinities. I really don't know where to even start with this one whatsoever. Any help? Or is it unsolvable?

Note that the non-transcendental irrational numbers form a subset of $\Bbb A = \overline {\Bbb Q} \subset \Bbb C$, the set of complex algebraic numbers (i.e. the algebraic closure of $\Bbb Q$).
From a purely set-theoretic perspective, let us show that $\Bbb A$ is countable. For $a_0, \dots, a_n \in \Bbb Q$, let $R(a_0, \dots, a_n)$ be the set of the roots of the polynomial $a_0 + \dots + a_n x^n$ - a finite set with at most $n$ distinct elements. Note that since every element of $\Bbb A$ is the root of a such polynomial, then
$$\Bbb A = \bigcup \limits _{n \ge 1} \bigcup \limits _{(a_0, \dots, a_n) \in \Bbb Q ^{n+1}} R(a_0, \dots, a_n) .$$
Since $\Bbb Q$ is countable, so will be $\Bbb Q ^{n+1} \space \forall n \ge 1$, so the inner union is a countable union of finite sets and therefore countable. Now, the outer union will be a countable union of countable sets, so again countable. The transcendental numbers will therefore be uncountable, therefore much more.
From a measure-theoretic point of view, countable sets are null sets for the Lebesgue measure, therefore again the set of the transcendental ones will be much larger.
Finally, from a topological point of view, it is slightly more technical to show that both types of numbers that you ask about are dense in $\Bbb C$, so from this point of view, they are equally many.
These statements remain true even when you consider only real algebraic and transcendental numbers.
