Can a finite, type $n+k$ spectrum be a (non-iterated) colimit of finite, type $n$ spectra for $k \geq 2$?

Question

By the thick subcategory theorem, if $X, Y$ are finite $p$-local spectra of type $m,n$ respectively, then $Y$ can be built from $Y$ in a finite number of "steps" iff $n \geq m$. Here, a "step" can be given by taking a cofiber, a direct sum, or a retract of previously constructed spectra.

Moreover, by the nilpotence theorem, it's always possible to pass from height $n$ to height $n+1$ in one step: if $X$ has type $n$, then it admits a $v_n$-self map, whose cofiber is of type $n+1$.

Question 1: Fix a prime $p$ and $n \in \mathbb N$. Does there exist a(n $\infty$-)category $J$, and an $\infty$-functor $F$ from $J$ to the category of type $n$ spectra, whose colimit $\varinjlim F$ exists and is a nonzero finite spectrum $Y$ of type $\geq n+2$?

Question 2: Same question, but we only require the nonzero finite spectrum $Y$ of type $\geq n+2$ to be a retract of $\varinjlim F$.

I think Question 2 is probably the better formulation. Note that by compactness, in (2) we may assume that $J$ is a finite ($\infty$-)category, and thus we may assume without loss of generality that $J = \Delta^{op}_{\leq N}$ is a truncated simplex category.

Answer 1

Yes, in a silly way : let $X,Y$ be nonzero, $X$ has type $n$ and $Y$ $n+2$.

Then $X\oplus Y$ has type $n$, and has $Y$ as a retract, so it solves Question 2 with $J= *$.

But in fact, taking the cofiber of $id_X\oplus 0$ on this spectrum yields $Y\oplus \Sigma Y$, which also has type $n+2$, so it solves Question 1 too.