Is there a forcing which only collapses cardinals when squared? Call a forcing notion $\mathbb{P}$ briefly tame if $\mathbb{P}$ preserves cardinals but $\mathbb{P}^2$ does not. I vaguely recall seeing an argument that such things can't exist, but I can't reconstruct it at the moment. My question is whether my recollection is just wrong:

Can briefly tame forcings exist in $\mathsf{ZFC}$?

Note that it's certainly the case that the square of a forcing notion can be much nastier than the original forcing itself. For example, there is a proper forcing whose square is non-proper - indeed, that forcing preserves $\omega_1$ but its square does not! Of course, it doesn't count since it does collapse $\omega_2$, but it is worryingly close to an example.
 A: This is an expansion of tzoorp's comment above, which observed that an existing answer to the question linked in the OP - due to Miha Habic - actually answers this one as well. I've made this answer CW to avoid reputation gain, and will delete it if either tzoorp or Miha add an answer of their own.

There is a forcing $\mathbb{P}$ with the following property: $\mathbb{P}$ does not collapse cardinals, but $\mathbb{P}^2$ sometimes collapses cardinals.
Specifically, working over a ground model with a Suslin tree $T$, let $\mathbb{P}_{0,T}$ be the forcing which specializes $T$ and let $\mathbb{P}_{1,T}$ be the forcing shooting a branch through $T$. These forcings are each c.c.c., hence do not collapse cardinals. However, their product does collapse $\omega_1$. Consequently, if we let $\mathbb{P}$ be their lottery sum, then $\mathbb{P}$ itself is c.c.c. but some $\mathbb{P}^2$ generics - namely, those whose left coordinate shoots a branch and whose right coordinate specializes, or vice versa - collapse $\omega_1$.
(And if we want we can make everything nicely definable by working over a ground model of $\mathsf{ZFC+V=L}$ and taking $T$ to be the $L$-least Suslin tree.)

This leaves open the strong form of the question: whether there is consistently a forcing which never collapses cardinals but whose square always collapses cardinals. The answer to that turns out to be yes.
