Solution:
We wish to find integers $a$, $z$ such that $48z \cdot 10^{a} + z = z\left(48 \cdot 10^{a} + 1\right)$ is a perfect square, where $z < 10^{a}$. This would prove that there exists a $48$-good perfect square because we are pulling off the last $a$ digits of the number and get two integers $48z$ and $z$.

To make $z$ small by keeping the product a perfect square, we'd like $48 \cdot 10^{a} + 1$ to be divisible by some reasonably large square. Take $a = 42 = \varphi(49)$. By Euler's theorem, $10^{42} \equiv 1 \pmod{49}$, so $48 \cdot 10^{a} + 1$ is a multiple of $49$.

Then we can take $z = \frac{48 \cdot 10^{a} + 1}{49}$. (Clearly $z < 10^{a}$, so we're fine.) Then we have
$$z\left(48 \cdot 10^{a} + 1\right) = \left(\frac{48 \cdot 10^{42} + 1}{7}\right)^{2}.$$