Solution:
If $\alpha$ and $\beta$ are both integers, then
$$[m \alpha]+[m \beta]=m \alpha+m \beta=m(\alpha+\beta)=m^{2}$$
This proves one implication.

Observe that $\alpha+\beta=m$ and $\alpha \beta=n$. We use the property of integer function: $x-1<[x] \leq x$ for any real number $x$. Thus
$$m^{2}-2=m(\alpha+\beta)-2=m \alpha-1+m \beta-1<[m \alpha]+[m \beta] \leq m(\alpha+\beta)=m^{2}.$$
Since $m$ and $n$ are positive integers, both $\alpha$ and $\beta$ must be positive. If $m \geq 2$, we observe that there is no square between $m^{2}-2$ and $m^{2}$. Hence, either $m=1$ or $[m \alpha]+[m \beta]=m^{2}$. If $m=1$, then $\alpha+\beta=1$ implies that both $\alpha$ and $\beta$ are positive reals smaller than 1. Hence $n=\alpha \beta$ cannot be a positive integer. We conclude that $[m \alpha]+[m \beta]=m^{2}$.

Putting $m=\alpha+\beta$ in this relation, we get
$$\left[\alpha^{2}+n\right]+\left[\beta^{2}+n\right]= (\alpha+\beta)^{2}$$
Using $[x+k]=[x]+k$ for any real number $x$ and integer $k$, this reduces to
$$\left[\alpha^{2}\right]+\left[\beta^{2}\right]=\alpha^{2}+\beta^{2}$$
This shows that $\alpha^{2}$ and $\beta^{2}$ are both integers. On the other hand,
$$\alpha^{2}-\beta^{2}=(\alpha+\beta)(\alpha-\beta)=m(\alpha-\beta)$$
Thus
$$(\alpha-\beta)=\frac{\alpha^{2}-\beta^{2}}{m}$$
is a rational number. Since $\alpha+\beta=m$ is a rational number, it follows that both $\alpha$ and $\beta$ are rational numbers. However, both $\alpha^{2}$ and $\beta^{2}$ are integers. Hence each of $\alpha$ and $\beta$ is an integer.