Suppose that $M$ is a simple (or irreducible) $R$-module. Does this imply that $M$ is projective?
Vector spaces are all projective since they are free, so no counter example can be gotten from there.
I tried to find a counter-example with the $\mathbb{Z}$-modules $\mathbb{Z}/{p\mathbb Z}$ where $p$ is prime (since they are irreducible) but could not get anywhere.
At this point I'm not even sure that the result is false. Please give only hints and not the answer.
Your attempt was really good! Since $\mathbb{Z}/p\mathbb{Z}$ are very good examples of such objects. You already know that they are irreducible. Now we just need to show that they are not projective. So assume that they are projective and consider the projection $$\pi:\mathbb{Z} \twoheadrightarrow \mathbb{Z}/p\mathbb{Z}$$ this is a projection and together with $\mathrm{id}:\mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z} $ this means that there needs to be some map $f:\mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}$ such that $\pi \circ f=\mathrm{id}$. But the only morphism $\mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}$ is the zero morphism which is a contradiction.