Let $\boldsymbol{v}_1=(x_1,y_1),\boldsymbol{v}_2=(x_2,y_2)\in\mathbb{R}^2$ be linearly independent over $\mathbb{R}$.
Define $\boldsymbol{v}\in\mathbb{R}^2$ to be transcendental if at least one of its coordinates is transcendental.
Does every $\mathbb{Z}$-basis of $\mathbb{Z}\boldsymbol{v}_1\oplus\mathbb{Z}\boldsymbol{v}_2$ have the same number of transcendental elements?
The example $\mathbb{Z}(\pi,1)\oplus\mathbb{Z}(\pi,0)=\mathbb{Z}(0,1)\oplus\mathbb{Z}(\pi,0)$ shows that the number of transcendental elements in a basis can change from $2$ to $1$.
It seems reasonable to expect that algebraicity be an invariant of $\mathbb{Z}$-bases of $\mathbb{Z}\boldsymbol{v}_1\oplus\mathbb{Z}\boldsymbol{v}_2$, given that $\boldsymbol{v}_1,\boldsymbol{v}_2\in\mathbb{R}^2$ are $\mathbb{R}$-linearly independent. So we change our definition...
Define $\boldsymbol{v}\in\mathbb{R}^2$ to be transcendental if $s(\boldsymbol{v})\neq\boldsymbol{0}$ for every monic irreducible $s(x)\in\mathbb{Z}[x]$ under componentwise addition and multiplication in $\mathbb{R}^2$. If $s(x)$ is irreducible, then $s(0)\neq 0$ so $s(a,0)\neq \boldsymbol{0}\neq s(0,a)$ for every $a\in\mathbb{R}$. Thus, $\boldsymbol{v}\in\mathbb{R}^2$ is transcendental if one of its coordinates is $0$. Also, $\boldsymbol{v}$ is transcendental if one of its coordinates is trascendental. With this background in place, let us ask the same question as before, but now using our new definition:
Question: Does every $\mathbb{Z}$-basis of $\mathbb{Z}\boldsymbol{v}_1\oplus\mathbb{Z}\boldsymbol{v}_2$ have the same number of transcendental elements?
As a number is transcendental iff every rational multiple is, we can as well work with $\mathbf Q$-span of $v_1,v_2$. Now traasition matrix for from given basis to another basis $w_1, w_2$ means we will have rational number as its entries. SO the $w_j$'s are rational linear combination of $v_1$ and $v_2$. This gives affirmative answer to your question.