Assuming you meant $r = \lambda^2$ (and also assuming you don't want repeated blocks), a proper approach might be to poke around the properties of $(r,\lambda)$-designs and their constructions to give necessary and sufficient conditions for the existence of such designs. But if you only need infinitely many nontrivial examples for various mixes of block sizes, you can construct them by generalizing the previous method as follows:

To obtain an $(r,\lambda)$-design with the desired property, you take a union of disjoint $S(2,k,v)$s like the previous method. But this time, you need more than one block size. I assume that you're familier with combinatorics and can prove the existence of set $\mathcal{S}$ of disjoint $S(2,k,v)$s with fairly large cardinality $\vert \mathcal{S} \vert$ on your own.

If you take a union of $x_i$ disjoint $S(2,k_i,v)$s, the resulting set forms the block set $\mathcal{B}$ of an $(r,\lambda)$-design with

$$r = \sum_i \frac{x_i(v-1)}{k_i-1}$$ and $$\lambda = \sum_i x_i.$$ So you only need to find suitable $x_i$ that satisfy $$\sum_i \frac{x_i(v-1)}{k_i-1} = \left(\sum_i x_i \right)^2$$ while keeping in mind that you should find $x_i$ disjoint $S(2,k_i,v)$s for each $i$. This way, in principle, you should be able to construct an $(r,\lambda)$-design with $r = \lambda^2$ and your favorite block size variation. I doubt this method covers all the parameters satisfying some necessary conditions you can derive by a simple counting argument, though. I didn't even do a back-of-the-envelop calculation.

Anyway, to illustrate how this construction works, consider the case when you want only block size, say, $3$ and $5$. Then the sums in the equation only have two terms each:

$$\frac{x_0(v-1)}{4}+\frac{x_1(v-1)}{2} = (x_0 + x_1)^2.$$

You need integer solutions while ensuring you sure have the specified numbers of disjoint guys. But it's not that difficult. For example, take $x_0 = \frac{2(v-1)}{9}$ and $x_1 = \frac{v-1}{9}$. I pulled these numbers off the top of my head, but this should be an example of solutions. You can check it or completely solve it for yourself. Anyway, all you have to do now is to see if there are $\frac{2(v-1)}{9}$ disjoint $S(2,5,v)$s and $\frac{v-1}{9}$ disjoint $S(2,3,v)$s. But this is almost trivially true because $x_0, x_1 \ll v$.

For instance, for $k=3$ you can use (part of) a large set of Steiner triple systems, which contain $v-2$ disjoint $S(2,3v)$s. For $k=5$, because you only need to find $\frac{2(v-1)}{9}$ disjoint $S(2,5,v)$s among the humongous numer ${{v}\choose{5}}$ of all $5$-tuples, you should be able to find them quite easily. For example, you can use the probabilistic method to show that if $v$ is sufficiently large, whatever $S(2,k,v)$ you have, you can find $\frac{2(v-1)}{9}$ isomorphisms (i.e., permutations of points) that lead to mutually disjoint designs. My quick calculation says you can easily find $v$ disjoint-inducing parmutations for $k \geq 5$, so gathering $\frac{2(v-1)}{9}$ disjoint $S(2,5,v)$s is no problem. Of course you can use whatever method you like to give disjoint designs. The point is that you pick small enough solutions $x_i$ (relative to $v$) so that you can find disjoint guys without a problem.

Assuming you meant $r = \lambda^2$ (and also assuming you don't want repeated blocks), a proper approach might be to poke around the properties of $(r,\lambda)$-designs and their constructions to give necessary and sufficient conditions for the existence of such designs. But if you only need infinitely many nontrivial examples for various mixes of block sizes, you can construct them by generalizing the previous method as follows:

To obtain an $(r,\lambda)$-design with the desired property, you take a union of disjoint $S(2,k,v)$s like the previous method. But this time, you need more than one block size. I assume that you're familier with combinatorics and can prove the existence of set $\mathcal{S}$ of disjoint $S(2,k,v)$s with fairly large cardinality $\vert \mathcal{S} \vert$ on your own.

If you take a union of $x_i$ disjoint $S(2,k_i,v)$s, the resulting set forms the block set $\mathcal{B}$ of an $(r,\lambda)$-design with

$$r = \sum_i \frac{x_i(v-1)}{k_i-1}$$ and $$\lambda = \sum_i x_i.$$ So you only need to find suitable $x_i$ that satisfy $$\sum_i \frac{x_i(v-1)}{k_i-1} = \left(\sum_i x_i \right)^2$$ while keeping in mind that you should find $x_i$ disjoint $S(2,k_i,v)$s for each $i$. This way, in principle, you should be able to construct an $(r,\lambda)$-design with $r = \lambda^2$ and your favorite block size variation. I doubt this method covers all the parameters satisfying some necessary conditions you can derive by a simple counting argument, though. I didn't even do a back-of-the-envelop calculation.

Anyway, to illustrate how this construction works, consider the case when you want only block size, say, $3$ and $5$. Then the sums in the equation only have two terms each:

$$\frac{x_0(v-1)}{4}+\frac{x_1(v-1)}{2} = (x_0 + x_1)^2.$$

You need integer solutions while ensuring you sure have the specified numbers of disjoint guys. But it's not that difficult. For example, take $x_0 = \frac{2(v-1)}{9}$ and $x_1 = \frac{v-1}{9}$. I pulled these numbers off the top of my head, but this should be an example of solutions. You can check it or solve the equation in a general way for yourself. Anyway, all you have to do now is to see if there are $\frac{2(v-1)}{9}$ disjoint $S(2,5,v)$s and $\frac{v-1}{9}$ disjoint $S(2,3,v)$s. But this is almost trivially true because $x_0, x_1 \ll v$.

For instance, for $k=3$ you can use (part of) a large set of Steiner triple systems, which contain $v-2$ disjoint $S(2,3v)$s. For $k=5$, because you only need to find $\frac{2(v-1)}{9}$ disjoint $S(2,5,v)$s among the humongous numer ${{v}\choose{5}}$ of all $5$-tuples, you should be able to find them quite easily. For example, you can use the probabilistic method to show that if $v$ is sufficiently large, whatever $S(2,k,v)$ you have, you can find $\frac{2(v-1)}{9}$ isomorphisms (i.e., permutations of points) that lead to mutually disjoint designs. My quick calculation says you can easily find $v$ disjoint-inducing parmutations for $k \geq 5$, so gathering $\frac{2(v-1)}{9}$ disjoint $S(2,5,v)$s is no problem. Of course you can use whatever method you like to give disjoint designs. The point is that you pick small enough solutions $x_i$ (relative to $v$) so that you can find disjoint guys without a problem.

Edit: I didn't mention, but this works only when $v$ satisfies sufficient conditions for the existence of $S(2,k_i,v)$ for all $i$. So, the above example with $k_0=3$ and $k_1=5$ is implicitly assuming $v \equiv 1, 3 \pmod{6}$ and $v \equiv 1, 5 \pmod{20}$ (and $v$ large enough).