A summary of textbook material on algebraic extensions of fields and algebraic elements (off the top of my head, just to get this question off the unanswered list).
Definition. If $K\subseteq L$ are fields and $z\in L$, we say that $z$ is algebraic over $K$, if there is a non-zero polynomial $p(x)\in K[x]$ such that $p(z)=0$. We say that $L$ is algebraic over $K$, if all the elements of $L$ are algebraic over $K$.
Basic observation. In the above setting $z$ is algebraic over $K$ if and only if there exists a vector space $V\subseteq L,V\neq\{0\}$, finite dimensional over $K$, such that $V$ is closed under multiplication by $z$.
Proof (sketch). If $z$ is a zero of a polynomial $p(x)\in K[x]$ of degree $n$, then $$V=K\cdot 1+K\cdot z+\cdots+K\cdot z^{n-1}$$ is closed under multiplication by $z$ because $z^n$ can be written as a $K$-linear combination of lower powers of $z$. For the other direction we can consider the characteristic polynomial $m(x)\in K[x]$ of the $K$-linear transformation $\rho:V\to V, \rho(x)=zx$. By basic linear algebra $m(z)=0$.
Corollary 1. If $[L:K]<\infty$ then $L$ is algebraic over $K$.
Proof. In this case we can use $V=L$ for all $z\in L$.
In particular because $\Bbb{Q}(i)$ is a 2-dimensional vector space over $\Bbb{Q}$, all its elements are algebraic over $\Bbb{Q}$.
Corollary. If $K\subseteq L\subseteq M$ are fields, $z\in M$ is algebraic over $L$, and $L$ is algebraic over $K$, then $z$ is algebraic over $K$.
Proof (Sketch). Let $p_L(x)=a_0+a_1x+\cdots a_nx^n\in L[x]$ be a non-zero polynomial such that $p_L(z)=0$. Because all the coefficients $a_i\in L$ there exist finite dimensional $(/K)$ subspaces $V_i,i=0,\ldots,n$, such that $V_i$
stable under multiplication by $a_i$. We then easily see that the space
$$
V=V_0\cdot V_1\cdots V_n\cdot W,
$$
where $W=K+K\cdot z+\cdots K\cdot z^n$, is stable under multiplication by $z$. Here I define the product $U_1\cdot U_2$ of two subspaces, $U_1$ and $U_2$, to be the space of finite $K$-linear combinations
$$
U_1U_2=\{\sum_{j=1}^m a_ju_{1j}u_{2j}\mid a_j\in K, u_{1j}\in U_1, u_{2j}\in U_2\}.
$$
We easily see that $\dim_K U_1\cdot U_2\le \dim_K U_1\cdot \dim_K U_2$. Consequently
$$
\dim_KV\le n\cdot\prod_{i=0}^n\dim_KV_i
$$
is finite. The claim follows.
Your claim is answered in the affirmative as a consequence of Corollary 2. Any number that is algebraic over $\Bbb{Q}(i)$ is necessarily also algebraic over $\Bbb{Q}$.
In your case an easier route to the destination is to observe that if
$$p(x)=a_0+a_1x+\cdots+a_nx^n$$
has coefficients in $\Bbb{Q}(i)$ (and $p(z)=0$ for some $z$), then, denoting
polynomial gotten by conjugating all the coefficients
$$
\overline{p}(x)=\overline{a_0}+\overline{a_1}x+\cdots+\overline{a_n}x^n,
$$
we see that the product $p(x)\overline{p}(x)$ has
- $z$ as a zero,
- its coefficients in $\Bbb{Q}(i)$, and
- its coefficients are real.
- Therefore its coefficients are actually rational.
A similar argument is available more generally when $L/K$ is a Galois extension, when we can apply all the field automorphisms to the coefficients of $p(x)\in L[x]$ and as the product get a polynomial with coefficients in the smaller field $K$.
