Real/Complex Zeros for Filter Transfer Function
Would the numerator of the transfer function then be equal to s−2

No, because S-2 implies Zero at $$(\sigma,\omega)=(2,0)$$ while you assumed Zero at $$\omega =2$$  and you can see in both cases imaginary parts are different.

The question states that there are multiple transmission zeros, but does not state how many zeros are at ω=2

only two possibilities are there either 1 Zero or 2 Zero at ω=2 , but two Zero at ω=2  is not possible because it leads to complex cofficient of numerator and which is not realizable and similarly two Zero at infinity is also not possible due to same reason as above.

I assumed that the zero at ω=2 was complex, leading to a conjugate pair (s+2j)(s−2j)

Your assumption seems correct let's see why-
Although Zero is given at ω=2 so one reasonable conclusion would be a zero at $$(\sigma,\omega)=(\sigma,2)$$ but it is also given that one Zero is infinity from which you can conclude that numerator is of 2nd degree polynomial . But for physically realizable filters one of the condition is that -all the coefficient of numerator and denominator should be real . Hence if one Zero is at $$S-(\sigma+j2) $$ so other would be at $$S-(\sigma-j2)$$.
if we consider real part of Zero as a variable ,we cannot calculate its value because we have only one condition(DC gain) and two variables , so the only logical interpretation of $$\omega =2$$ is that its Zero is at (0,2) hence numerator will be $$(S-j2)(S+j2)=S^2+4$$