Check $m^2 - m + 11$ modulo $2,$ $3,$ $5,$ and $7:$ it is never divisible by any of them

So every prime factor is at least $11.$ Test the candidates $11^4$ and $11^3 \cdot 13,$ using $m \equiv 0$ or $1 \pmod{11}.$

Let $e(m) = m^2 - m + 11.$ Since $m^2 - m$ is always even, $e(m)$ is odd. Checking all residues shows $m^2 - m + 11$ is never $0$ modulo $3,$ $5,$ or $7$ either, so every prime factor of $e(m)$ is at least $11.$ A product of four such primes is at least $11^4 = 14641,$ and the two smallest candidates are $11^4$ and $11^3 \cdot 13 = 17303.$

For $e(m) = 14641,$ the discriminant of $m^2 - m - 14630 = 0$ is $58521,$ which lies strictly between $241^2 = 58081$ and $242^2 = 58564,$ so there is no integer solution. For $e(m) = 17303:$ since $e(m) = m(m - 1) + 11$ must be divisible by $11,$ either $m = 11k$ or $m = 11k + 1.$ Trying $m = 11k$ gives $11k^2 - k + 1 = 1573,$ that is $k(11k - 1) = 1572,$ which $k = 12$ satisfies: $12 \cdot 131 = 1572.$

Since $e$ is increasing for $m \ge 1,$ every smaller $m$ has $e(m) \lt 17303,$ and the only four-prime value below that, $11^4,$ is unattainable. Hence the least $m$ is $11 \cdot 12 = 132,$ where $e(132) = 17303 = 11^3 \cdot 13.$