The classical proof of the primitive element theorem (over $\mathbb Q$) implies the stronger result that if $\alpha,\beta$ are two algebraic numbers over $\mathbb Q$, then ${\mathbb Q}(\alpha+t\beta)={\mathbb Q}(\alpha,\beta)$ for all but finitely many $t\in{\mathbb Q}$.
Conversely, given a finite $T\subseteq {\mathbb Q}$, is there a pair $(\alpha,\beta)$ such that ${\mathbb Q}(\alpha+t\beta) \neq {\mathbb Q}(\alpha,\beta)$ for each $t\in T~?$
I'm especially interested in the case $T=\lbrace 0,1,2,\ldots, n\rbrace$ for $n\in{\mathbb N}$.
My thoughts : it is easy to see that all the ${\mathbb Q}(\alpha+t\beta)$ for $t\in T$ must be distinct subfields of ${\mathbb Q}(\alpha,\beta)$, so the extension ${\mathbb Q}(\alpha,\beta)/{\mathbb Q}$ has at least $|T|$ intermediary fields. Its degree $d$ must therefore satisfy $2^{d!} \geq |T|$ (because the decomposition field has a degree $g$ such that $g\leq d!$, and there are at most $2^{g}$ subgroups (in fact, at most $2^{g}$ subsets) of the Galois group).
To put (a very minor improvement on) Camilo Arosemena's comment into a full answer : if we write $T=\lbrace t_1<t_2<\ldots<t_n\rbrace$, then we can take
$$ \begin{array}{lcl} \alpha &=& -(t_1\sqrt{p_1}+t_2\sqrt{p_2}+\ldots+t_n\sqrt{p_n}),\\ \beta &=& \sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n}+\sqrt{p_{n+1}} \end{array} $$
where the $p_i$ are distinct primes. Then ${\mathbb Q}(\alpha+t_k\beta)$ does not contain $\sqrt{p_k}$ whence ${\mathbb Q}(\alpha+t_k\beta) \neq {\mathbb Q}(\alpha,\beta)$.