Since $x^5-x-1$ is irreducible over $\mathbb{Q}$, then $[\mathbb{Q}(\alpha) \colon \mathbb{Q}] = 5$. Let $\beta \in \mathbb{Q}(\alpha)$ such that $f(\beta) = 0$, with $\deg(f) = 3$.

Suppose that $f$ is irreducible. Then $[\mathbb{Q}(\beta) \colon \mathbb{Q}] = 3$. But $\beta \in \mathbb{Q}(\alpha)$, so we have $[\mathbb{Q}(\alpha) \colon \mathbb{Q}(\beta)] \cdot [\mathbb{Q}(\beta) \colon \mathbb{Q}] = 5$, impossible since $[\mathbb{Q}(\beta) \colon \mathbb{Q}] = 3$ and $3$ does not divide $5$, a contradiction. Then $f$ is not irreducible, and since it has degree $3$ it must have a linear factor (it splits in either $3$ irreducibles of degree $1$, or one of degree $1$ and one of degree $2$).

In general, it's not enough to have $m,n$ coprime, we need $(n, m!) = 1$. Suppose you have a degree $10$ extension and a degree $7$ polynomial $f$ that splits in a factor of degree $2$ and a factor of degree $5$. Then $f$ might have roots in the extended field that are not in $\mathbb{Q}$, e.g. take $f(x) = (x^2+1)(x^5-x-1)$ and $\mathbb{Q}(\alpha, i)$. The extension has degree $10$ but $f$ has $i$ as a root, which is in the extension but not in $\mathbb{Q}$.

However, if $(n,m!) =1$ then the same argument as in the first part ensures that there cannot be any root in the extension that is not already in $\mathbb{Q}$.