How would I go about proving

$$ a \equiv b\pmod{m}$$
given

$$ 
a \equiv b\pmod{m^2} 
$$

I suspected this was false, so I tried to find a counterexample using a [program](https://gist.github.com/MayorMonty/d15ad34cd5e7558da37c0910a4fecefb), and there does not appear to be a counterexample for $m$ in (2, 16).

What I have tried so far:
$$ 
a \equiv b\pmod{m^2} 
$$
$$
a = b + m^2k
$$
$$
a - b = m^2k
$$
$$
\sqrt{a - b} = \sqrt{m^2k}
$$

Which would imply that $\sqrt{a-b} \equiv 0\pmod{m}$ assuming that $k$ is a perfect square. It feels like I'm barking up the wrong tree, however, and I'm not sure where to go from here.