Let $l = \text{lcm}(m, n)$ and write $n = l/u, m = l/v$. Then
$$\alpha^{1/n} = (\alpha^u)^{1/l}, \quad \beta^{1/m} = (\beta^v)^{1/l}.$$
Note that $\alpha^u$ and $\beta^v$ are also rationals. Thus it is sufficient to consider the case $n = m$. Let us write $a = \alpha^{1/n}$, $b = \beta^{1/m}$ and $c = a + b$. We are given that $c$ is rational.

Lemma 1: Let $a, b$ be real numbers such that $a + b = c$ is a rational number. Let $f(x)$ and $g(x)$ be the irreducible polynomials of $a$ and $b$ over $\mathbb{Q}$. Then $\deg(f(x)) = \deg(g(x))$.

Proof of Lemma 1: Let $p = \deg(f(x))$ and $q = \deg(g(x))$. Then
$$f(x) = x^p + a_{p-1}x^{p-1} + \dots + a_0, \quad g(x) = x^q + b_{q-1}x^{q-1} + \dots + b_0$$
and these are polynomials with rational coefficients. Since $f(a) = 0$, we have $f(c-b) = 0$. Thus $b$ satisfies the polynomial $f(c-x)$ over $\mathbb{Q}$. Since $g(x)$ is the irreducible polynomial of $b$ over $\mathbb{Q}$, it follows that $g(x)$ divides $f(c-x)$. Hence $q \le p$. Similarly we show that $p \le q$ and we obtain $p = q$.

We prove the result by induction on $n$. If $n = 1$, then $\alpha, \beta$ are rational by the given condition. If $n = 2$, then
$$\alpha^{1/2} - \beta^{1/2} = \frac{\alpha - \beta}{\alpha^{1/2} - \beta^{1/2}}$$
which shows that $\alpha^{1/2} - \beta^{1/2}$ is also rational. Combined with the given condition that $\alpha^{1/2} + \beta^{1/2}$ is rational, it follows that each of $\alpha^{1/2}, \beta^{1/2}$ is rational.

Suppose the result is true for $k = 0, 1, 2, \dots, n-1$. Let $f(x)$ be the irreducible polynomial of $a = \alpha^{1/n}$ over $\mathbb{Q}$ and $g(x)$ be that of $b = \beta^{1/n}$ over $\mathbb{Q}$. Then the lemma 1 shows that $\deg(f(x)) = \deg(g(x)) = m$, say. But we know that $a$ is a root of $x^n - \alpha = 0$, $b$ is a root of $x^n - \beta = 0$ and $b = c - a$. Thus $(c-a)^n = b^n = \beta$. Hence $(a-c)^n = (-1)^n \beta$. This shows that $a$ is a root of $(x-c)^n - (-1)^n \beta = 0$. This is a polynomial with rational coefficients. Thus $a$ is a root of
$$h(x) = (x - c)^n - (-1)^n \beta - (x^n - \alpha).$$
Now $\deg(h(x)) = n-1$, it follows that $\deg(f(x)) \le n-1$. Thus we have $m \le n-1 < n$. Let $\omega$ be a primitive $n$-th root of unity. Then
$$x^n - \alpha = \prod_{j=0}^{n-1} (x - a\omega^j).$$
Since $f(x)$ is the irreducible polynomial of $a$ over $\mathbb{Q}$, and $a$ is a root of $x^n - \alpha$, we see that $f(x)$ divides $x^n - \alpha$. Hence $x^n - \alpha = f(x)q(x)$ for some rational polynomial $q(x)$. Now the factors of $f(x)$ are all of the form $(x - a\omega^j)$. Hence the constant coefficient of $f(x)$ which is a rational number must be of the form $\pm a^m$, where $m = \deg(f(x))$. Since $m < n$, we see that $a^m \in \mathbb{Q}$ for some $m < n$. Similarly, we see that $b^m$ is also in $\mathbb{Q}$. Since $m < n$, induction hypothesis shows that $a, b$ are also in $\mathbb{Q}$. This completes the proof.