It is well-known that there does necessarily not exist a $\kappa$-saturated elementary extension of cardinality $\kappa$ of a given structure (if, for instance, if the structure is unstable.)
What if we weaken the requirement from saturation to compactness? Here, $M$ being $\kappa$-compact means that it realizes all partial type of size less than $\kappa$.
It seems that there still seem to be too many partial types of the required size, so there may not exist such a thing. Is there a reference on this matter?
You note correctly that $\kappa$-compact is a weaker notion than $\kappa$-saturated (for infinite $\kappa$).
Proof: Suppose $\Sigma$ is a partial type with $|\Sigma|<\kappa$. Then if $A$ is the set of parameters mentioned by $\Sigma$, we have $|A|<\kappa$. So we can extend $\Sigma$ to a complete type $p$ over $A$, and any model realizing $p$ realizes $\Sigma$. So any $\kappa$-saturated model is $\kappa$-compact. $\square$
The converse is true when $\kappa>|L|$ (by $|L|$, I mean the number of formulas with no parameters).
Proof: Suppose $p$ is a complete type over $A$, with $|A|<\kappa$. Then $|p| = \max(|L|,|A|)<\kappa$, and any $\kappa$-compact model realizes $p$. So any $\kappa$-compact model is $\kappa$-saturated. $\square$
So if $\kappa>|L|$ is such that $M$ has no $\kappa$-saturated elementary extension of size $\kappa$, then $M$ also has no $\kappa$-compact elementary extension of size $\kappa$!
OK, what about when $\kappa \leq |L|$?