A model $M$ is decidable if the set of all first-order formulas which are true in $M$ is a recursive set. And a model is $1$-decidable if the set of all existential formulas which are true in $M$ is a recursive set. This paper gives an example of a linear order which is $1$-decidable (in fact $n$-decidable for all $n$) but not decidable. As pointed out by Laurent Moret-Bailly in the comments, by this MO answer, the field $\mathbb{R}(t)$ is 1-decidable but not decidable.

My question is, are there any known examples of algebraic extensions of $\mathbb{Q}$ which are $1$-decidable but not decidable?