It is clear that $\mathbb{Q}(\alpha+i) \subseteq \mathbb{Q}(\alpha, i)$. You could try to show that $\alpha\in\mathbb{Q}(\alpha+i)$. Then $i\in\mathbb{Q}(\alpha+i)$ and so $\mathbb{Q}(\alpha,i)\subseteq\mathbb{Q}(\alpha+i)$ thereby solving the question.
Normally, you would try to find some expression involving only addition, subtraction, multiplication, division, rational numbers and $\alpha+i$ that equals $\alpha$ (or $i$, as Joel Cohen did). There is also an algorithmic way to check whether $\alpha\in\mathbb{Q}(\alpha+i)$. It is basically a structured way of finding such an expression.
Write down $(\alpha+i)^k$ in coordinates with respect to the basis you found. For example, if $k=3$, you get $ (\alpha+i)^3 = \alpha^3+3i\alpha^2 - 3\alpha - i = 3i\alpha^2-i-1-2\alpha$ where I have used the relation $\alpha^3-\alpha+1=0$. Let us (for clarity) write this as a 6-tuple of coordinates with respect to $\{1,\alpha,\alpha^2,i,i\alpha,i\alpha^2\}$. We get $ (\alpha+i)^3 \cong (-1,-2,0,-1,0,3). $ Doing this not only for $k=3$, but for $k\in\{0,\ldots,5\}$, you get six tuples of coordinates. The question whether $\alpha\in\mathbb{Q}(\alpha+i)$ then comes down to: is there a linear combination of these tuples giving $(0,1,0,0,0,0)\cong\alpha$?
If you find such a linear combination (in this case you will, because the six vectors will be linearly independent hence span all of $\mathbb{Q}^6$), you have found a linear combination of powers of $(\alpha+i)$ (i.e. a polynomial expression in $(\alpha+i)$) that equals $\alpha$. This of course guarantees that $\alpha\in\mathbb{Q}(\alpha+i)$, and you're done. I admit this takes some calculations, but it is a surefire way of solving the question.
Remark. There is a nice link with a common proof of the primitive element theorem. It is proved by finding $\lambda$ such that $\mathbb{Q}(a,b)=\mathbb{Q}(a+\lambda b)$. In the proof, it turns out that this will be true for all but finitely many values for $\lambda$, so typically we have that $\mathbb{Q}(a,b)=\mathbb{Q}(a+\lambda b)$. This is the same as saying that $n$ different vectors in $\mathbb{Q}^n$ will typically span all of $\mathbb{Q}^n$. I have never checked whether this link can be made precise, so it's just the intuition I have about this.