degree of a sum of two algebraic numbers

I claim the order is $2019^2$, but I'm not going to prove some of my sub-claims (and I'm not 100% certain they're all true). So it's not quite a full answer, but maybe it's a new way for you to look at the problem.

Throughout, let $\zeta$ denote a primitive $2019^{\tiny\mbox{th}}$ root of unity, let $\alpha = \sqrt[2019]{2}$, and let $\beta = \sqrt[2019]{3}$.

The first sub-claim is that $\mathbb{Q}(\alpha, \beta)$ has degree $2019^2$ over $\mathbb{Q}$. Let $f(x) = x^{2019}-2$ and $g(x)=x^{2019}-3$. If $g(x)$ splits over $\mathbb{Q}(\alpha)$, then, since all roots of $g$ in $\mathbb{C}$ have the form $\zeta^k \beta$, then $\beta \in \mathbb{Q}(\alpha)$ (which I sub-claim to be false without proof), as the constant term of any factor of $g$ will be of the form $\zeta^\ell \beta$, which is not real unless $\ell = 0 \pmod{2019}$, in which case $\beta \in \mathbb{Q}(\alpha)$. So $g(x)$ is the minimal polynomial of $\beta$ over $\mathbb{Q}(\alpha)$, and thus $\mathbb{Q}(\alpha,\beta)$ has degree $2019^2$ over $\mathbb{Q}$.

The second sub-claim is that $\alpha+\beta$ generates $\mathbb{Q}(\alpha,\beta)$ over $\mathbb{Q}$, i.e., is a primitive element. This can be done by showing that $$\frac{(\zeta^m - 1)\alpha}{(1-\zeta^n)\beta} \ne 1,$$ for any choice of $m,n$ (which I sub-claim to be true without proof). To see why, see the constructive proof of the primitive element theorem. This implies that $\mathbb{Q}(\alpha+\beta)=\mathbb{Q}(\alpha,\beta)$ has degree $2019^2$ over $\mathbb{Q}$.

Note, as @ÍgjøgnumMeg indicates in the comments, there is an algorithm for finding the minimal polynomial of $\alpha + \beta$. Since it lies in $\mathbb{Q}(\alpha,\beta)$, which has basis $\{\alpha^i\beta^j : 0\le i,j\le 2018\}$ (this hinges upon my first sub-claim), and so you can raise $\alpha+\beta$ to higher and higher powers, writing these powers in terms of the basis, and look for linear dependency among the thus-far-computed powers until there is one.