Are there more transcendental numbers or irrational numbers that are not transcendental?

The non-transcendental numbers (otherwise known as the algebraic numbers – Wikipedia link) comprise a countably infinite set, whereas the transcendental numbers are uncountably infinite.

(Why are there only countably many algebraic numbers? Because we can group them according to what polynomial in $\mathbb{Q}[x]$ they are a root of, and any such polynomial has finitely many roots, and there are only countably many such polynomials.)

The point is: in colloquial terms, there are more transcendental numbers than algebraic numbers.

Therefore, there are certainly more transcendental numbers than there are algebraic numbers that also are not rational.

The set of algebraic numbers $\mathbb A$ is countable, so $\mathbb A\cap (\mathbb R\setminus \mathbb Q)$ is also countable. On the other hand, the set of transcendental numbers $\mathbb R\setminus\mathbb A$ must be uncountable, so $$|\mathbb R\setminus\mathbb A|>|\mathbb A\cap(\mathbb R\setminus\mathbb Q)|.$$

Hint: the set of algebraic numbers is Countable.

For proving this, you can show that the polynomials with integer coefficients form a countable set, and that the set of their roots is also countable. The latter is just the Set of Algebraic Numbers (over $\mathbb{C}$). As the Algebraic Numbers over $\mathbb{R}$ form a subset of the former set, it must also be Countable.