Using the parametrization $ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} (R+\cos u)\cos v \\ (R+\cos u)\sin v \\ \sin u \end{bmatrix} $ for $u,v \in [0,2\pi]$, we get that the first fundamental form is $ ds^2 = du^2 + (R+\cos u)^2 \, dv^2. $ Changing the metric conformally by a factor of $(R+\cos u)^{-1}$ we get a new metric tensor $ \frac{ds^2}{(R+\cos u)^2} = \frac{du^2}{(R+\cos u)^2} + dv^2.$ This metric (as any metric tensor of the form $E(u)\, du^2 + G(v) \, dv^2$) has zero Gaussian curvature, so it is locally isometric to the plane. Furthermore, the isometry is easy to write down in terms of the parameters as $ \Phi(u,v) = \left(\int_0^u \frac{dt}{R+\cos t}, v\right). $ The image of this torus under $\phi$ is the flat torus $[0,a] \times [0,b]$ with the usual identifications, where $b=2\pi$ and $ a = \int_0^{2\pi} \frac{dt}{R+\cos t} = \frac{2\pi}{\sqrt{R^2-1}} $ where the evaluation of the integral is a standard application of the residue theorem.
Let $\mathbb{T}_a$ be the flat torus generated by gluing together opposite sides of $[0,1] \times [0,a]$, and let $\mathbb{T}'_R$ be the embedded torus described above with parameter $R>1$, and let $\approx$ denote conformal equivalence. As outlined in this post, $\mathbb{T}_a \approx \mathbb{T}_b$ iff $a=b$ or $a=1/b$, i.e., if the fundamental rectangles are similar. Now from the explicit conformal map given above, we know $\mathbb{T}'_R \approx \mathbb{T}_{\sqrt{R^2-1}}$.
Summary
(1) Two such embedded tori with parameters $R,S>1$ are conformally equivalent iff $R=S \quad \text{ or }\quad (R^2-1)(S^2-1) = 1.$
(2) Two such flat tori with parameters $a,b>0$ are conformally equivalent iff $a=b \quad \text{ or }\quad ab=1.$
(3) An embedded torus with parameter $R>1$ and a flat torus with parameter $a>0$ are conformally equivalent iff $a=\sqrt{R^2-1} \quad \text{ or }\quad a\sqrt{R^2-1} = 1.$