at.algebraic topology - Are there pairs of highly connected finite CW-complexes with the same homotopy groups?

Fix an integer n. Can you find two finite CW-complexes X and Y which

* are both n connected,

* are not homotopy equivalent, yet

* $pi_q X approx pi_q Y$ for all $q$.

In Are there two non-homotopy equivalent spaces with equal homotopy groups? some solutions are given with n=0 or 1. Along the same lines, you can get an example with n=3, as follows. If $Fto Eto B$ is a fiber bundle of connected spaces such that the inclusion $Fto E$ is null homotopic, then there is a weak equivalence $Omega Bapprox Ftimes Omega E$. Thus two such fibrations with the same $F$ and $E$ have base spaces with isomorphic homotopy groups.

Let $E=S^{4m-1}times S^{4n-1}$. Think of the spheres as unit spheres in the quaternionic vector spaces $mathbb{H}^m, mathbb{H}^n$, so that the group of unit quarternions $S^3subset mathbb{H}$ acts freely on both. Quotienting out by the action on one factor or another, we get fibrations
$$ S^3 to E to mathbb{HP}^{m-1} times S^{4n-1},qquad S^3to Eto S^{4m-1}times mathbb{HP}^{n-1}.$$
The inclusions of the fibers are null homotopic if $m,n>1$, so the base spaces have the same homotopy groups and are 3-connected, but aren't homotopy equivalent if $mneq n$.

There aren't any n-connected lie groups (or even finite loop spaces) for $ngeq 3$, so you can't push this trick any further.

Is there any way to approach this problem? Or reduce it to some well-known hard problem?

(Note: the finiteness condition is crucial; without it, you can easily build examples using fibrations of Eilenberg-MacLane spaces, for instance.)