It is not possible with 7 (i.e., with a triangle $T$ and a quadrilateral $Q$). I write a rough proof.
First, any quadrilateral $Q$ lying in a plane $\pi$ can be partitioned in two triangles $Q_1$ and $Q_2$, whose common edge is a diagonal $d$ of $Q$. Now the intersection of the triangle $T$ with $\pi$ consists of two points (otherwise they are coplanar and the conclusion is trivial). Since the polygons are not linked, there are two cases: either both points lie outside $Q$, or they both lie inside $Q$.
Case 1: both lie outside. An inspection of the cases shows that we can just translate $T$ in direction parallel to one of the bisectors of the triangles $Q_1$ or $Q_2$ starting from one of the vertices not belonging to $d$.
Case 2: both lie inside. We will consider a modified problem: given a quadrilateral $Q$ and two initial points $x(0),y(0)$ inside it, find two continuous curves $x(t),y(t)$ such that the distance between the two is decreasing in time to 0, and they never lie in $Q$. The solution is, e.g., a linear homotopy that sends both points to the midpoint of $d$. Now we only have to realize $x(t)$ and $y(t)$ as the intersections of $T$ and the plane $\pi$. One can convince themselves that this is possible by pulling $T$ in direction orthogonal to $\pi$ while translating it.