I have to find the convergence domain $\mathcal{D}$ of the Taylor series of $f(z) =\sin (z_1 + z_2^2)$ at the origin. I claim $\mathcal{D}=\mathbb{C}^2$ and I prove this in the following manner:
We have $$\sin (z_1 + z_2^2) = \sin(z_1)\cos(z_2^2) + \sin(z_2^2)\cos(z_1)$$ Since the taylor series of each multiplicand of each summand converge everywhere (as a single variable complex function), so does $f(z)$, therefore the claim is proved.
Is my proof/argument correct?
Just like in one variable, the Taylor series around $p$ will converge on any polydisc centered at $p$ that fits in the domain on which the function is holomorphic. The way to prove it is just the iterated Cauchy integral formula leading to the several variable Cauchy estimates. See e.g. https://www.jirka.org/scv/scv.pdf sections 1.1 and 1.2, and in particular Theorem 1.2.1. Since your function is holomorphic in ${\mathbb C}^2$ and so any polydisc fits in your domain, and so the series centered at any point will converge in the entire ${\mathbb C}^2$. Your argument would also work as long as you can use that the product and addition of power series has at least that same domain of convergence, which is not as easy to prove from scratch, but it is elementary.