To unpack and correct Dan's answer:

The extensions $\mathrm{Ext}^1(A,B)$ for $A$ and $B$ mixed Hodge structures are given by considering the direct sum $C_{\mathbb{Q}}=A_{\mathbb{Q}}\oplus B_{\mathbb{Q}}$ with the obvious weight filtration $W_mC_{\mathbb{Q}}=W_mA_{\mathbb{Q}}\oplus W_mB_{\mathbb{Q}}$ , and choosing the Hodge filtration $F_kC=\{(a,b+\varphi(a))\in C_{\mathbb{C}}|a\in A_{\mathbb{C}},b\in B_{\mathbb{C}}\}$ for $\varphi \colon A\to B$ a map preserving the weight filtration. The resulting extension is trivial if and only if $\varphi=\varphi'+\varphi''$ is the sum of a map $\varphi'$ preserving the Hodge filtration, and a map $\varphi''\colon A_{\mathbb{Q}}\to B_{\mathbb{Q}}$ which is defined rationally (if you want integral or real MHS, just replace $\mathbb{Q}$ by $\mathbb{Z}$ or $\mathbb{R}$ in the definition of $\varphi''$.

In the specific case of $\mathrm{Ext}^1(\mathbb{Q}(n),\mathbb{Q}(m))$, we have:

- if $m<n$, then only the trivial map preserves the weight filtration, so the Ext group is 0.
- if $m=n$, then any map preserves the weight filtration, but it also preserves the Hodge filtration, and the extensions are trivial.
- if $m> n$, then any map preserves the weight filtration, but only the trivial preserves the Hodge filtration. So we end up with $\mathrm{Ext}^1(\mathbb{Q}(n),\mathbb{Q}(m))=\mathrm{Hom}_{\mathbb C}(\mathbb C,\mathbb C)/\mathrm{Hom}_{\mathbb Q}(\mathbb Q,\mathbb Q)$, so it's isomorphic to $\mathbb{C}/\mathbb{Q}$, though I think people like to normalize to $\mathbb{C}/(2\pi i)^{m-n}\mathbb{Q}$ (this just depends on whether you choose the generator of $\mathrm{Hom}_{\mathbb C}(\mathbb C,\mathbb C)$ to preserve the rational structures and thus send $(2\pi i)^n$ to $(2\pi i)^m$, or be the identity map).

Dan's answer was assuming you want integral Hodge structures, so you get $\mathbb{C}/\mathbb{Z}\cong \mathbb{C}^*$ instead.