So I read your edit, and here's the answer: Yes. Infinitely many of them exist.

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.

Edit2: The $2$-designs constructed here have a pair of parallel blocks unless you pick too small $v$. For example, the $S(2,3,7)$ is the only nontrivial guy that doesn't have parallel blocks. So larger v always works for your problem with the added condition if $k$ is $3$. I'm on my iPhone, so allow me to omit the details; it's almost trivial if you think about if you can construct a Steiner $2$-design without parallel blocks.

Edit3: Now I'm back on my mac. Here's why every $S(2,k,v)$ for large enough $v$ has a pair of blocks that don't intersect each other:

Let $(V, \mathcal{B})$ be an $S(2,k,v)$ in which no pair of blocks are parallel. Take a block $B = \lbrace x, y , z\rbrace \in \mathcal{B}$. Because every block must intersect $B$, the degree of $x$ (i.e., the number of blocks that contain $x$) is $\frac{\vert \mathcal{B}\vert-1}{k} +1 = \frac{v(v-1)-k(k-1)}{k^2(k-1)}+1$. However, because we have $r = \frac{v-1}{k-1}$, we must have $\frac{v-1}{k-1} = \frac{v(v-1)-k(k-1)}{k^2(k-1)}+1$. Because the left-hand side is $O(v)$ while the right-hand side is $O(v^2)$, this can hold only when $v$ is small enough compared to $k$. (You can actually solve it to express $v$ by $k$ and check when this condition can be met if you want.) So, for fixed $k$, if you take a large $S(2,k,v)$, it must have a parallel block. Thus, by taking $\frac{v-1}{k-1}$ copies or pasting $\frac{v-1}{k-1}$ mutually disjoint ones, you get the desired $2$-design as long as $v$ is large enough.