So I read your edit, and here's the answer: Yes. Infinitely many of them exist. But the construction is so obvious that I still kind of feel like you have other conditions.
Anyway, if you only need a $2$-design with $r = \lambda^2$ which has at least one pair of blocks intersecting each other, then you can simply copy a Steiner $2$-design. Take an $S(2,k,v)$ (i.e., a Steiner $2$-design of order $v$ and block size $k$, which exist for infinitely many pairs of $v$ and $k$). Then its repetition number $r$ is exactly $r = \frac{v-1}{k-1}$. Since it's a Steiner $2$-design, its index $\lambda$ is one. If you make a copy of this design $r = \frac{v-1}{k-1}$ times, then what you get is a $2$-design of order $v$, block size $k$, index $\frac{v-1}{k-1}$, and repetition number $(\frac{v-1}{k-1})^2$.
To make the above trivial method even more trivial, here's an example: Take the Fano plane. It's the unique $S(2,3,7)$. Copy this guy $3$ times. Then you get the $2$-$(7,3,3)$ design you mentioned in your question. Why we copied $3$ times is because it's $\frac{v-1}{k-1}=\frac{6}{2}=3$. So, for example, because an $S(2,3,v)$ exists for all $v \equiv 1, 3 \pmod{6}$, you can have infinitely many examples of what you want by pasting the same $S(2,3,v)$ $\frac{v-1}{2}$ times.
Edit: I don't know if this makes a difference, but if you don't want repeated blocks, there are still infinitely many examples. A $2$-design is simple if it has no repeated blocks (i.e., all blocks are distinct). Two $2$-designs $D_0$ and $D_1$ on the same point set are disjoint if they have no common block.
Now, if you have $\frac{v-1}{k-1}$ mutually disjoint $S(2,k,v)$s, taking the union of their block sets will give you a simple $2$-design of the same order and the same block size which satisfies $r = \lambda^2$. Since you didn't specify the block size, you can use, for example, the large set of Steiner triple systems of order $v$, which is a set of $v-2$ mutually disjoint $S(2,3,v)$s. (A large set is a set of disjoint designs of block size $k$ in which the union of block sets is the set of all $k$-subset of the point set. For the case of Steiner triple systems, you have $v-2$ mutually disjoint $S(2,3,v)$s.) A large set of $S(2,3,v)s$ exist for all possible $v \not= 7$. Since $\frac{v-1}{2} < v-2$ for all $v > 3$, you can surely have $\frac{v-1}{2}$ mutually disjoint $S(2,3,v)$s. Hence, infinitely many no-repeated-block versions also exist.