I want to split a message using Shamir's Secret Sharing by splitting each byte of it so the field I'm using is $2^8$. Assume $(2,2)$ scheme, i.e., splitting it into $2$ shares and both of them are needed for reconstruction. By having one share, can I guess the second (I have max $256$ options) and try to build the polynomial to reveal the secret? am I missing something here? so larger $k$ is important here? Thanks!
There is a one-to-one correspondence between the value of the second share and the value of the secret: each of the $2^8$ possible values of the second share will give a (distinct) value of the secret among the $2^8$ possible values for it, and vice versa. Hence, guessing the second share is exactly the same thing as guessing the secret.