, 2012

Answer: $(n; m) = (4; 2)$.
By condition,
$$n^2 + n + 1 = (m^2 + n - 3)(m^2 - n + 5) = m^4 + m^2 + 8m - 15.$$
Consider the obtained equation
$$n^2 + n - (m^4 + m^2 + 8m - 16) = 0 \quad (1)$$
as a quadratic equation with respect to $n$. It has positive integer roots only if the determinant $D = 4m^4+4m^2+32m-63$ of this equation is a perfect square of some integer number. But
$$D = 4m^4 + 4m^2 + 32m - 63 = (2m^2 + 2)^2 - 4(m - 4)^2 - 3 < (2m^2 + 2)^2$$
for any natural number $m$, and
$$D = 4m^4 + 4m^2 + 32m - 63 = (2m^2 + 1)^2 + 32(m - 2) > (2m^2 + 1)^2$$
for any natural number $m > 2$. Therefore, (1) has the natural roots only if $m = 1$ or $m = 2$.
If $m = 1$, then $n^2 + n + 6 = 0$, so either $n = -2$ or $n \in \mathbb{Z}$.
If $m = 2$, then $n^2 + n - 20 = 0$, so either $n = -5$ or $n = 4$.
Thus, (4; 2) is a unique pair of positive integers satisfying the problem condition.