Finding every $n$ such that $n\times$ ('reverse' number of $n)=m^2$ such as $1584\times 4851={2772}^2$

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Let $r(n)$ be the 'reverse' number of $n$ in the decimal system. For example, $r(1234)=4321$.

Then, here is my question.

Question : Can we find every $n(\in\mathbb N)$, which is not a square number, such that
$$n\times r(n)=\text{a square number ?} $$

Motivation : First, I got interested in the followings : $$169\times 961={403}^2, 144\times 441={252}^2$$ Note that these examples are not what we want because each of $n$ is a square number. These got me interested in the above question. Then, I've already found the followings by using computer :

$$1584\times 4851={2772}^2$$ $$15984\times 48951={27972}^2$$

Then, I realized that these are the special cases of the following equation :

$$\begin{align}1599\cdots 984\times 4899\cdots 951={2799\cdots 972}^2\qquad(\star)\end{align}$$ Note that each of $99\cdots 9$ has $m$ consective $9$s where $m$ is a non-negative integer.

We can see that $(\star)$ is true because $$16\cdot({10}^{m+2}-1)\times 49\cdot({10}^{m+2}-1)=\{28\cdot({10}^{m+2}-1)\}^2.$$

However, I don't have any good idea for finding every possible $n$. Can anyone help?