Trying to think of different ways to prove this, besides method of sqrt. Help will be greatly appreciated.
$b \mid m^2$ and $b \mid n^2$, so $b^2 \mid m^2 n^2$, so $b \mid mn$.
To see why $a^2 \mid b^2$ implies $a \mid b$, see the link already provided by Martin in his comment to the question.
Hint $\rm\ mn/b\ $ is a root of $\rm\: x^2 - ac\:$ so is integral by the Rational Root Test.
You can use prime factorizations of integers... You know that $ab^2c = m^2n^2 = (mn)^2$. If a prime $p$ appears to some power $e$ in the prime factorization of $(mn)^2$, then it appears to the power $e$ in $ab^2c$, so it appears to a power $d \leq e$ in the prime factorization of $b^2$. This is equivalent to the statement that if $p$ appears to the power $e' = e/2$ in the prime factorization of $mn$, then it appears to the power $d' = d/2\leq e'$ in the prime factorization of $b$.
Thus every prime appears to at least a large a power in the prime factorization of $mn$ as it does in the prime factorization of $b$. So $b$ divides $mn$.
$ab=m^2$, $bc=n^2$ so $acb^2=(mn)^2$ and $b^2$ divides $(mn)^2$.
Thus $mn/b$ is a rational number whose square is an integer. By unique factorization into primes it now follows that $mn/b$ is an integer; any prime factor in the denominator of the lowest form of a rational number will also occur in the factorization of the denominator of its square.