The answer is $k = 3$. Take $x=12, y = 18$ and eliminate the small cases, and expand to prove that $k = 3$ is indeed the answer.

$ x\mid y^{2} ,\ y\mid z^{2} ,z\mid x^{2} \ \ \cdots ( 1)\\ xyz\mid ( x+y+z)^{l} \ \ \cdots ( 2) $ Note that $ ( x,y,z) =\left( 2^{4} ,2^{2} ,2\right)$ satisfies $ ( 1)$. From $ ( 2)$, we have $ 2^{7} \mid \left( 2^{3} +2 +1\right) 2^{l}$, which implise $ l\geq 7$. $ ( 1)$ hold if and only if $ \nu _{p}( x) \leq 2\nu _{p}( y)$, $ \nu _{p}( y) \leq 2\nu _{p}( z)$, and $ \nu _{p}( z) \leq 2\nu _{p}( x)$ for all primes $ p$. $ ( 2)$ hold if and only if $ \nu _{p}( x) +\nu _{p}( y) +\nu _{p}( z) \leq l\cdot \nu _{p}( x+y+z)$ for all primes $ p$. Let $ p$ be an arbitrary prime number. WLOG, $ \min\{\nu _{p}( x) ,\nu _{p}( y) ,\nu _{p}( z)\} =\nu _{p}( z)$. Then we have $ \nu _{p}( x+y+z) \geq \nu _{p}( z)$. From $ \nu _{p}( x) \leq 2\nu _{p}( y)$ and $ \nu _{p}( y) \leq 2\nu _{p}( z)$, we have $ \nu _{p}( x) +\nu _{p}( y) \leq 6\nu _{p}( z)$. So we have $ \nu _{p}( x) +\nu _{p}( y) +\nu _{p}( z) \leq 7\nu _{p}( z)$. Therefore, the answer is $ \boxed{l=7}$.