Here is the kind of method I had in mind.

We have the elliptic curve Kummer sequence
$$0 \to E[n] \to E \to E \to 0,$$
Here I denote by $E[n]$ the $n$-torsion group scheme of $E$. Applying Galois cohomology we obtain
$$0 \to E(\mathbb{Q})/nE(\mathbb{Q}) \to H^1(\mathbb{Q}, E[n]) \to H^1(\mathbb{Q}, E)[n] \to 0.$$

By the Mordell-Weil theorem, the group $E(\mathbb{Q})/nE(\mathbb{Q})$ is *finite* (its cardinality grows roughly like $n^{\mathrm{rank}(E)}$).

Thus it suffices to show that $H^1(\mathbb{Q}, E[n])$ is infinite. I think that this should be some general property of Galois cohomology for non-trivial finite abelian group schemes over number fields, which probably you already know about.

Anyway, the argument should go as follows: Choose a splitting field $k/\mathbb{Q}$ for $E[n]$. We then apply inflation-restriction to obtain
$$0 \to H^1(\mathrm{Gal}(k/\mathbb{Q}), E[n]) \to H^1(\mathbb{Q}, E[n]) \to H^1(k, (\mathbb{Z}/n\mathbb{Z})^2)^{\mathrm{Gal}(k/\mathbb{Q})} \to H^2(\mathrm{Gal}(k/\mathbb{Q}), E[n]).$$
The first and the latter group are finite. The group $H^1(k, (\mathbb{Z}/n\mathbb{Z})^2)$ is clearly infinite, and I think that it is still infinite after taking Galois invariants. Though this last step is the part I did not fully check. Is it clear to you?