Let $A, B$ be Morita equivalent $\mathbb{E}_1$-ring spectra. Fix an an $(A, B)$-bimodule $P$ and a $(B, A)$-bimodule $Q$ such that $P \otimes_B Q \cong A$ and $Q \otimes_A P \cong B$. If $A$ is bounded below then $B$ is bounded below, because $P, Q$ are perfect $A$-modules (hence bounded below) and \begin{align*} \operatorname{conn}(B) &= \operatorname{conn}(Q \otimes_A P) \\ &\geq \operatorname{conn}(Q \otimes P) \\ &\geq \operatorname{conn}(Q) + \operatorname{conn}(P) + 1 \\ &> -\infty. \end{align*} 

Is it true that if $A$ is connective then $B$ is connective? Ie, can a connective and nonconnective ring spectrum be Morita equivalent? This is of course *not* possible for $\mathbb{E}_\infty$-ring spectra.

One thing I was thinking about is that $\operatorname{End}_B(P) \cong A$ is connective, and if $\pi_*(P)$ is a faithful $\pi_*(B)$-module this would imply $B$ must be connective as well. It seems plausible for it to be faithful in some sense because it's invertible. But I'm not if the "some sense" would persist after taking homotopy groups.