Solution:

We will show that the only pair satisfying the conditions of the problem is $\{1,2\}$. (Clearly there is also a solution if $a=b=n$ for every $n$, but in this case $\{a, b\}$ is not a pair; whether or not this case has been considered has no bearing on the grading anyway).
We may of course assume, by symmetry, that $a>b$.

For $n \geq 2$, using the fact that $x_{1} \ldots x_{n-1}=x_{n}-5$, we can also write $x_{n+1}=\left(x_{1} \ldots x_{n-1}\right) x_{n}+5=\left(x_{n}-5\right) x_{n}+5=x_{n}^{2}-5 x_{n}+5$.
Observe that for $a>b$ we have $x_{a}=x_{1} \ldots x_{b} \ldots x_{a-1}+5$, so, if a prime divides both $x_{a}$ and $x_{b}$, then it divides $x_{a}-x_{1} \ldots x_{b} \ldots x_{a-1}=5$.
However, 5 divides $x_{n+1}=x_{n}^{2}-5 x_{n}+5$ if and only if it divides $x_{n}$: it follows that if a term of the sequence is not divisible by 5, neither is the following one. Since $x_{1}$ is not divisible by 5, no term of the sequence is divisible by 5.
This shows that $x_{a}$ and $x_{b}$ never have common prime divisors, so $x_{a} x_{b}$ is a perfect square if and only if both $x_{a}$ and $x_{b}$ are: we therefore look for which perfect squares occur in the sequence.
Certainly $x_{1}=4$ and $x_{2}=9$ are squares, so $\{1,2\}$ is a solution.
On the other hand, since $x_{2}=9$, for $n \geq 2$ the number $x_{n}$ is of the form $x_{n}=9 k+5$ and hence, in particular, leaves remainder 2 upon division by 3. Now it is easy to check that no square has this property, since the square of a multiple of 3 is a multiple of 3 while the square of a number of the form $3 m \pm 1$ is $9 m^{2} \pm 6 m+1$ and hence leaves remainder 1 upon division by 3. Hence there are no other solutions besides the one found.

Alternatively, in a perhaps more elementary form, since for $n \geq 2$ we have $x_{n+1}=x_{n}^{2}-5 x_{n}+5$, we can study the equation $x^{2}-5 x+5=a^{2}$ with $x, a$ positive integers.
Multiplying both sides by 4 we obtain
$$\begin{gathered} 4 a^{2}=4 x^{2}-20 x+20=(2 x-5)^{2}-5 \Rightarrow 5=(2 x-5)^{2}-4 a^{2} \\ \Rightarrow(2 x-2 a-5)(2 x+2 a-5)=5 . \end{gathered}$$
Since 5 is only divisible by $\pm 1, \pm 5$ and $2 x+a-5>2 x-a-5$, there are only the two possibilities $2 x+2 a-5=-1, 2 x-2 a-5=-5$ and $2 x+2 a-5=5, 2 x-2 a-5=1$, which lead respectively to $x=1, a=1$ and $x=4, a=1$. The number 1, however, cannot appear in the sequence, because $x_{1}=4$ and each term is larger than the previous one.
We deduce again that no perfect squares appear in the sequence, apart from $x_{1}=4, x_{2}=9$ (which are not expressed in the form $x^{2}-5 x+5$).