(Solution by V. Vityaz.) First, $q \neq r$ otherwise from
$$q(q^2 - q - 1) = r(2r + 3) \quad (1)$$
we would have $r^2 - r - 1 = 2r + 3$ which gives $r = 4$, which is not prime.

Hence (1) implies $q^2 - q - 1 = kr$, so $q^2 - q - 1 = kr$, $2r + 3 = kq$. Eliminating $r$ we obtain
$$2q^2 - (2 + k^2)q + 3k - 2 = 0. \quad (2)$$
The discriminant of this equation is equal to $D = k^4 + 4k^2 - 24k + 20$ and $D$ must be a perfect square. Note that $(k^2)^2 < D < (k^2 + 2)^2$ for $k > 5$. Hence
$$D = (k^2 + 1)^2 \Longleftrightarrow 2k^2 - 24k + 19 = 0.$$
But the last equation has no integer roots.

If $k = 1$, then (2) becomes $2k^2 - 3q + 1 = 0$, $q = 1$ which is not prime. Note that $k$ is odd since $kr = q^2 - q - 1$ is odd. If $k = 3$, then (2) has no integer solutions. If $k = 5$, then (2) becomes $2q^2 - 27q + 13 = 0$, then $q = 13$, $r = 31$, and $n = r(2r + 3) = 2015$.