Are all maximally symmetric spacetimes conformally flat? What about the converse?

If I'm not mistaken, one of the properties of maximally symmetric spacetimes is that the Riemann tensor can be written as $$R_{abcd} = \frac{R}{d(d-1)}(g_{ac}g_{bd} - g_{ad}g_{bc})$$, which would imply that the Weyl tensor vanishes.

And Wikipedia states that a d-dimensional spacetime with $$d \geq 4$$ is conformally flat if and only if the Weyl tensor vanishes.

This is telling me that the answer to my first question is "Yes", but what about the second one? I guess that one could have a spacetime with vanishing Weyl tensor but with a Riemann tensor that's not only "trace" but also has "symmetric" components coming from $$R_{ij} \neq \frac{1}{d} R g_{ab}$$. Are there any easy examples of such spacetimes?

The usual examples given for both concepts (MS and CF) are Minkowski, AdS, and dS; and so I am a bit confused.

Thank you!

All FLRW spacetimes are conformally flat, and they are not maximally symmetric unless $$ρ=-p=\text{constant}$$ (in which case they are Minkowski or (anti) de Sitter).