What's the fuss about $m$ ?

Let $x = \log_2 m$ , then the equation can be written as $x^2 + 3x - 10 = 0.$ The sum of the zeroes of a quadratic is the negative of the linear coefficient, so $x_1 + x_2 = -3.$ The product of the corresponding $m$ values is $m_1 \cdot m_2 = 2^{x_1 + x_2} = 2^{-3} = \boxed{0.125}.$ Note that it is not even necessary to solve for the individual values of $m$ or $x$ ...

$\begin{aligned} (\log_{2}{m})^2+\log_{2}{m^3} = & 10 \\ \implies (\log_{2}{m})^2+3\log_{2}{m}-10=& 0 \quad \quad \quad [\log_{a}{b^{c}}=c\log_{a}{b}] \\ \implies (a-2)(a+5)=& 0 \quad \quad \quad [\text{Let}~ \log_{2}{m}~ \text{be}~ a] \\ \implies a=& 2,-5 \implies \log_{2}{m}=2,-5 \\ \therefore m=& 4,\dfrac{1}{32} \end{aligned}$