Solution:

Multiplying the two equations gives $$zw + 12i + 20i - \frac{240}{zw} = (5 + i)(-4 + 10i) = -30 + 46i,$$ so $zw - \frac{240}{zw} = -30 + 14i.$ Setting $v = zw$ yields $v^2 + (30 - 14i)v - 240 = 0.$

By the quadratic formula, $v = -15 + 7i \pm \sqrt{(15 - 7i)^2 + 240} = -15 + 7i \pm \sqrt{416 - 210i}.$ Writing $(a + bi)^2 = 416 - 210i$ requires $a^2 - b^2 = 416$ and $ab = -105,$ which gives $a + bi = \pm(21 - 5i).$ Hence $v = 6 + 2i$ or $v = -36 + 12i,$ with $|v|^2 = 40$ or $1440.$

The smaller value is attained: $z = 1 - i,$ $w = 2 + 4i$ satisfies both equations with $zw = 6 + 2i.$ So the smallest possible value of $|zw|^2$ is $40.$