$\sf ZFC + Classes$ is a bi-sorted theory with lower cases standing for sets and upper cases for Classes; axioms include all $\sf ZFC - Extensionality$ axioms written in lower case, and the following axioms on top of bi-sorted ID-theory $^\dagger$:
Sorting: $\forall x \exists Y: x=Y$
Membership: $\forall X \ ( \exists Y( X \in Y) \leftrightarrow \exists z( z=X))$
Extensionality: $\forall X \forall Y (\forall z (z \in X \leftrightarrow z \in Y) \to X=Y)$
Comprehension: if $\phi(y)$ is a formula not using symbol $``X"$, and in which $y$ occur, with parameters (free variables other than $y$) among $\vec{A}$, then: $$\forall \vec{A} \, (\exists X \, \forall y \, (y \in X \leftrightarrow \phi(y)))$$
Define: $X=\{y \mid \phi\} \iff \forall y \, (y \in X \leftrightarrow \phi)$
Define: $\operatorname {set}(X) \iff \exists y: y=X$
Define: $\operatorname {proper class} (X) \iff \neg \operatorname {set}(X)$
Now $\sf ZFC + Classes$ is a conservative extension of $\sf ZFC$. However, it differs from $\sf NBG$ in that the class comprehension schema is impredicative, i.e., like $\sf MK$, it allows quantification over class variables. Also, it differs in that it doesn't prove global choice.
Is $\sf ZFC + Classes$ finitely axiomatizable?
$^\dagger$ bi-sorted ID theory is the extension of the logical axioms of bi-sorted $\sf FOL$ with the following axioms:
Reflexivity. $\forall x: x = x \\ \forall X: X=X$
Substitution for functions. For all variables $x,X,y,Y$ and any function symbol $f$:$$x = y \to f(..., x, ...) = f(..., y, ...) \\ x = Y \to f(..., x, ...) = f(..., Y, ...) \\ X = Y \to f(..., X, ...) = f(..., Y, ...)$$ Substitution for formulas. For any variables $x,X,y,Y$ and any formulas $\phi(x), \phi(X)$, if $\phi(x | y), \phi(x|Y) $ are formulas obtained by replacing any number of free occurrences of $x$ in $\phi(x)$ with $y,Y$ respectively, such that these remain free occurrences of $y,Y$ respectively; and if $\phi(X|y), \phi(X|Y)$ are obtained by replacing any number of free occurrences of $X$ in $\phi(X)$ with $y,Y$ respectively, such that these remain free occurrences of $y,Y$ respectively, then $$x = y \to (\phi(x) \to \phi(x|y)) \\x = Y \to (\phi(x) \to \phi(x|Y)) \\X = y \to (\phi(X) \to \phi(X|y))\\ X = Y \to (\phi(X) \to \phi(X|Y))$$.