Given a field extension $K/F$, the degree of the extension $K/F$ is defined as the dimension of $K$ viewed as a vectorial space over $F$.
In other words $$[K:F]=\dim_FK.$$
For your question, I let you to prove (if you haven't already proved it) that if $\alpha$ is algebraic over $F$, then $$F(\alpha)=\{g(\alpha): g(x)\in F[X]\; \text{and}\; g\equiv 0\; \text{or}\; \deg(g)<\deg(m_{\alpha,F})\}.$$
Now let's set $m_{\alpha,F}=p$, then if $p(x)=a_0x^n+\cdots +a_{n-1}x+a_n$ we claim that $\{1,\alpha,\ldots \alpha^{n-1}\}$ is a basis for $F(\alpha)$ over $F$. Indeed, every element of $F(\alpha)$ has the form $b_0\alpha^{k}+\cdots +b_{n-1}\alpha+b_n$ for some $ g(x)=b_0x^k+\cdots +b_{n-1} x+b_n\in F[x] $ with $k<n$, so it follows easily that $\{1,\alpha,\ldots \alpha^{n-1}\}$ is a generating set of $F(\alpha)$ as a vectorial space over $F$. To prove that the above set is l.i. we write $$c_0+c_1\alpha+\ldots c_{n-1}\alpha^{n-1}=0.$$
Then $q(x)=c_0+c_1x+\ldots c_{n-1}x^{n-1}\in F[x]$ satisfies $q(\alpha)=0$, so either $q\equiv 0$ or $p(x)\mid q(x)$. But the later is impossible because that would mean that $n=\deg(p)\le deg(q)=n-1$, an absurd. Therefore $q\equiv 0$, which implies that $c_0=c_1=\ldots =c_{n-1}=0$.
Hence $\{1,\alpha,\ldots \alpha^{n-1}\}$ is l.i. and thus it's a basis for $F(\alpha)$ over $F$. From this is immediate that $$[F(\alpha):F]=n=\deg(p)=\deg(m_{\alpha,F}).$$