Suppose $F$ is a field, $\alpha$ and $\beta$ are algebraic over $F$, $[F(\alpha,\beta):F] = mn$ for positive integers $m$ and $n$, and $[F(\alpha):F] = m$. Must there exist a $\gamma \in F(\alpha,\beta)$ for which $\gamma \notin F(\alpha)$, $[F(\gamma):F] = n$, and $F(\alpha,\gamma) = F(\alpha,\beta)$?
This came up when looking for the Galois group of the splitting field of $x^4 - 14x^2 +9$ over the rationals Q. The roots are $\pm \sqrt{7 \pm 2 \sqrt{10}}$, and I have convinced myself that the splitting field is $K = Q(\sqrt{7 + 2 \sqrt{10}},\sqrt{7 - 2 \sqrt{10}})$, an extension of $Q$ of degree 8. I was looking for a more convenient set of generators for describing the automorphisms of $K$, but was having trouble finding what I wanted, which led to the above question. That is, I was looking for a $\gamma \in Q(\sqrt{7 + 2 \sqrt{10}},\sqrt{7 - 2 \sqrt{10}})$ such that both $\gamma \notin Q(\sqrt{7 + 2 \sqrt{10}})$ and $[Q(\gamma):Q] =2$. I thought this would be useful so I could describe the 8 automorphisms of $K$ as the maps $\sqrt{7 + 2 \sqrt{10}} \mapsto \pm \sqrt{7 \pm 2 \sqrt{10}}$ and $\gamma \mapsto \pm \gamma$.
There are many counterexamples.
Perhaps the simplest: take $f(x)$ to be an irreducible polynomial with Galois group $S_m$ for $m > 4$, and let $n = m-1$. Let $\alpha$ and $\beta$ be two roots of $f(x)$. Then $F = \mathbf{Q}(\alpha,\beta)$ has degree $m(m-1) = mn$, and $\mathbf{Q}(\alpha)$ has degree $m$. But the splitting field of $f(x)$ doesn't even contain a subfield of degree $n=m-1$, since that would give an index $m-1$ subgroup of $S_m$ which does not exist for $m > 4$.
Another counterexample: take $n=m=2$, and $\alpha = \sqrt{2}$ and $\beta = \sqrt[4]{2}$.
More generally, if $F$ is any field of degree $mn$ with a subfield of degree $m$, then, by the primitive element theorem, you can write $E = \mathbf{Q}(\alpha)$ and $F = \mathbf{Q}(\beta) = \mathbf{Q}(\alpha,\beta)$, so your claim would imply that any field of degree $mn$ with a subfield of degree $m$ is the compositum of two fields of degree $m$ and $n$. Using Galois theory, you can then translate this into a property of subgroups of the Galois group of the splitting field $G$ (which I won't write down here) which need not hold for general $G$.