Note that the sum of the digits of this number is equal to
$$S(X_n) = 1 \cdot n + 5 \cdot (n - 1) + 6 = 6n + 1.$$
Further,
$$\begin{aligned} X_n &= \frac{10^{2n} - 1}{9} + \frac{4 \cdot (10^n - 1)}{9} + 1 \\ &= \frac{10^{2n} - 1 + 4 \cdot (10^n - 1) + 9}{9} \\ &= \frac{10^{2n} + 4 \cdot 10^n + 4}{9} \\ &= \left( \frac{10^n + 2}{3} \right)^2 \end{aligned}$$
is a perfect square. (Since $S(10^n + 2) = 3$, we have $10^n + 2 \vdots 3$.) It suffices to note that if $n = 6m^2 + 2m$ ($m \in \mathbb{N}$), then $S(X_n) = 6n + 1 = 6(6m^2 + 2m) + 1 = (6m + 1)^2$. Therefore, the numbers $X_{6m^2+2m}$ satisfy the problem condition for all $m \in \mathbb{N}$, and we have infinitely many such numbers.

Remark. Also we can consider
$$(2 \cdot 10^n + 1)^2$$