Answers by Alex Becker and Robert Mastragostino describe another statement:
$P_3:\exists x(S(x)\land \exists y(S(y)\land (x\neq y) \land \forall z(L(z) \to C(x, z)\to C(y, z))))$
where $S(x)$ means “$x$ is a student” and $L(x)$ means “$x$ is a class”. I suppose that your logic is 1-sorted, i. e., variables range over classes and students.
I denote your statement 2 as $P_2$. We cannot prove $P_2$ and $P_3$ equivalent. Suppose that there are 2 classes and no students. Then $P_2$ is true: take those classes as $x, y$; $x\neq y$ is true; for all $z$, $C(x, z)$ is false because it makes no sense for a class to be enrolled in something. $P_3$ is false because we can not choose $x$ because there are no students.
The correct translation of $P_2$ is “there are $x$ and $y$ such that for every $z$, $x$ and $y$ are distinct and if $x$ is enrolled in $z$ then $y$ is enrolled in $z$”. You must not tell words “student” and “class” because there is no $S$ and $L$ in $P_2$.
If you want to learn more, I recommend to consider a logic with sorts $S$ (students) and $L$ (classes). In this logic there are similar statements:
$P_4: (\exists x\in S(\exists y\in S(\forall z\in L((x\neq y) \land C(x, z)\to C(y, z)))))$
$P_5: (\exists x\in S(\exists y\in S((x\neq y) \land \forall z\in L(C(x, z)\to C(y, z)))))$
Answers by Alex Becker and Robert Mastragostino describe $P_5$. Note that $(x\neq y)$ sits in different places in $P_4$ and $P_5$. We cannot prove $P_4$ and $P_5$ equivalent. Suppose that there are only 1 student and no classes. Then $P_5$ is false and $P_4$ is true.