I was emailed the following argument:
We prove that $k[[x]]\otimes_{k} k((x))$ is not noetherian by showing directly that $k((x)) \otimes k((x))$ is not noetherian (as suggested by Georges Elencwajg). I will just handle the case $k=\bar{\mathbb{Q}}$ and then make a remark about the general case at the end.
The field $k((x))$ has only countably many finite separable extensions because every such extension is obtained by adjoining a root of $x$. On the other hand, the transcendence degree of $k((x))$ over $k$ must be uncountable because $k((x))$ is uncountable and $k$ is countable. Fix a transcendence basis $( t_i )_{i \in I}$ for $k((x))$ over $k$.
The algebraic extension $k((x))$ of $K := k((t_i))$ is algebraic with infinite separable degree. Indeed, if the separable degree was finite, then $K$ would admit at most countably many finite separable extensions as this is true for $k((x))$. This is absurd because $( t_i )$ is uncountable.
Because the separable degree of $k((x))$ over $k((t_i))$ is infinite, $(k((x)) \otimes_{K} k((x)))_{\text{red}}$ has infinitely many idempotents and so
$$k((x)) \otimes_{K} k((x))$$ and hence $$k((x)) \otimes_{k} k((x))$$
are non-noetherian. This completes the proof.
With work, this proof can be modified to hold when $k$ is a finite field $\mathbb{F}$. In this case, one must argue more carefully to show that $k((x))$ has only countably many finite separable extensions. (The email indicated that one should use local compactness together with Krasner's Lemma.) Finally, one can deduce the case of a more general field $k$ from the case $k=\bar{\mathbb{Q}}$ or $\mathbb{F}$ by using a faithfully flat descent argument.