Take a number $x=\overline{a_1a_2a_3\cdots a_n}$. Its repeated form is like $\overline{\underbrace{a_1a_2a_3\cdots a_n}_{\text{x}}\underbrace{a_1a_2a_3\cdots a_n}_{\text{x}}}$ And $\exists$ $s$ such that the repeated form is a perfect square, that is, $\overline{\underbrace{a_1a_2a_3\cdots a_n}_{\text{x}}\underbrace{a_1a_2a_3\cdots a_n}_{\text{x}}}=s^2$. In fact, I have done it and that's a pretty common problem.
Still, here's my solution:
Solution for $r=2$ (just a super gist): Let the number be $n$ with $m$ digits.. Its repeated form is $n(10^m+1)$ and we have to show that there exists $k$ such that $(10^m+1)n=k^2$. Since $n$ has $m$ digits and $10^m+1$ has $m+1$ digits, $n<10^n+1$. If GCD$(10^m+1,n)=1$, we will easily get a contradiction as $10^m+1$ has to be individually a square and that's not possible. Similarly, we will encounter a few more cases (I won't prove them as they are just easy and simple parity checking) and finally, we can claim that $10^m+1=a^2b$ and $n=bc^2$ where $a,b,c > 1$. We can check by Wolfram Alpha or simply make a program (C++, Python perhaps) and ensure that in case of $2^{11}+1$ we get such $a,b$ where $a=11, b=826446281$. Solve it to get $4\leq c\leq10$. Let's show that it holds for infinitely many. Just take any $\text{odd }x$ and it will hold for $10^{mx}+1$ as $10^m+1\mid 10^{mx}+1$ since $m,x$ are $\text{odd numbers}$. The value of $c$ will change. That's it!!
But what if the repeat is done $r$ times, that is, the repeated number is $\overline{\underbrace{a_1a_2a_3\cdots a_n}_{\text{1st set}}\underbrace{a_1a_2a_3\cdots a_n}_{\text{2nd set}}\cdots \underbrace{a_1a_2a_3\cdots a_n}_{r\text{-th set}}}$
In that case, do there exists a perfect square?
My take on this may not be pleasing.
So if we have a repeated square of the form $\underbrace{\overline{a_1 \ldots a_n}}_{\text{One time}} \underbrace{\overline{a_1 \ldots a_n}}_{\text{Two times}} \ldots \underbrace{\overline{a_1 \ldots a_n}}_{r \text{ times}}$.
First of all, this function has to be divisible by $10\ldots 01 0 \ldots 0 1 \ldots 01$ for some appropriate number of dots and repetitions.
The trick with $r=2$ was the following: Suppose that $100\ldots001$ has a square factor, for example $100000000001$ is a factor of $11^2 = 121$, along with $23 * 4093*8779$.
If we try to find the smallest number, that when multiplied with $100000000001$ gives a perfect square, then we can see that this number is $23 * 4093*8779$. Multiplying these two, we get the number $82644628100826446281$. This multiplied by $16$ gives $1322314049613223140496$, and multiplication by $25$, $36$ etc. gives all the other repeated squares I had mentioned in my comment earlier.
The important thing here, was the restriction of the number of digits that the number being multiplied with $100000000001$ had. It must be exactly of $11$ digits so that we can get a square upon multiplication.
Now, imagine doing a similar procedure for $r=3$. This means that we have to find ,for the number $100...0001000...0001$, a number with two conditions:
1) The number must have less than half the number of digits in $100...0001000...0001$ (so that on multiplying (along with multiplying by a suitable square factor to get exact repetition), you get a perfect square).
2) The number must be divisible by all the factors, which are contained in odd multiplicity in $100...0001000...0001$ (so that it is capable of completing the square).
Do you see how difficult this condition is to satisfy? Not only must the factors of odd multiplicity be small, their product must have half the digits of $100...0001000...0001$. This is not easy to satisfy at all, once you realize that most of these numbers end up having large prime factors, not of multiplicity two, killing the second condition entirely.
For $r \geq 4$, it's even worse: now, the product of the factors has to be less than one-third the size of $100...0001000...0001000...0001$. This is humungously difficult to satisfy, because one cannot predict the large prime factors.
Further, in the case of such repeated things being cubes or fourth powers, it's even more insane, since then we will be looking at factors not having multiplicity three or four respectively, (as we are then trying to multiply by a number to get a third or fourth power), and these will be extremely rare.
So, while I cannot rule out any example, I can safely say that empirically, I would not expect solutions to the repeated number squares to come easily for $r \geq 3$.