Perfect squares formed by two perfect squares like $49$ and $169$.

Let $$x_k = 2^{2k-1} 5^{2k} - 10^k + 1 = 4\underbrace{99...9}_{k-1}\underbrace{00...0}_{k-1}1.$$ Then $x_k^2$ has the desired property, as it is the concatenation of the squares of $$2^{k-1}5^k-1=4\underbrace{99...9}_{k-1} \quad \mbox{ and }\quad 10^k-1 = \underbrace{99...9}_{k}.$$

For small $k$, we get the following:

$$ k=1: \quad x_1^2 = 41^2 = \underbrace{16}_{4^2}\,\underbrace{81}_{9^2}. $$ $$ k=2: \quad x_2^2 = 4901^2 = \underbrace{2401}_{49^2}\,\underbrace{9801}_{99^2}. $$ $$ k=3: \quad x_3^2 = 499001^2 = \underbrace{249001}_{499^2}\,\underbrace{998001}_{999^2}. $$ $$ \vdots $$

It is an easy computation to check that this works for every $k$. Note that this method does not find all solutions, but it does show that there are infinitely many of them. Probably there will be other infinite families of solutions.