Edit: After rereading your question, it seems that you just want one surface $Y$ which is not birationally equivelant to a surface with picard rank $1$. This is a bit easier. It is not clear if you want $Y$ to have Picard rank $1$, but in both cases there are examples. For an example with Picard rank $1$ take $\mathbb{CP}^{2}$, for an example without picard rank $1$ take $\mathbb{P}^{1} \times E$ where $E$ is an elliptic curve. Below I show that there can be surfaces with picard rank $1$ which are not birational.

Yes, for example fake projective planes https://en.wikipedia.org/wiki/Fake_projective_plane. A fake projective is a smooth projective surface with the same Betti numbers as $\mathbb{CP}^{2}$.

They necessarily have Kodaira dimension $2$, which is different from the Kodaira dimension $\mathbb{CP}^{2}$ i.e. $-\infty$. This is not so hard to see, from some classical classification results on algebraic surfaces. A projective variety with $b_{2}=1$ is either Fano, Calabi-Yau or general type (just consider whether $K_{X} \in H^{2}(X,\mathbb{R})$ is positive, 0, or negative). Fano surfaces (also called del Pezzo surfaces) are classified into 10 topological types and have no fake projective planes. Calabi-Yau surfaces with $b_{1}=0$ have just two topological types, K3 and Enqriques surfaces, having $b_{2}=22$ and $b_{2}=10$ respectively.

A fake projective plane $X$ has Picard rank $1$, since Picard rank is bounded above by $b_{2}(X)=1=b_{2}(\mathbb{CP}^{2})$ by definition and is positive because there is an ample line bundle.

Let $X$ and $Y$ be smooth complex projective surface. If there exists a birational map $f: X \dashrightarrow Y$ which is an isomorphism between big open subsets of $X$ and $Y$, I say that $X$ and $X$ are birational equivalent.It is generally not such a good idea to give new definitions to terms that are already in common use. $\endgroup$