Let U be Maximal among non-principal ideals of R. (Question regarding phrasing) With regard the above statement, is the author trying to say:
(1):$ $ $\forall z \lhd R(z$ is non-principal $\land$ $U\subseteq z$ $\land$ $U\neq R$ $\Rightarrow U=z$ $ \lor$ $ z=R )$
or 
(2): Suppose U is non-principal. Then $ $ $\forall z \lhd R(z$ is non-principal $\land$ $U\subseteq z$ $\land$ $U\neq R$ $\Rightarrow U=z$ $ \lor$ $ z=R )$
If it is the first case, then there could be a possibility that a ring exists with a principal ideal - $U=(k)$ - such that all non-principal ideals are a strict subset of it. In which case 1 still holds since the premise is always false.

You should interpret it as "$U$ is a non-principal ideal with the property that there does not exist a non-principal ideal of $R$ strictly containing $U$."

If I were forced to encode encrypt what you said in in logical notation, I would write what those words say:
$U \text{ is non-principal }\wedge(\not\exists z\lhd R((z\text{ non-principal })\wedge (z\supsetneq U)))$
which would be equivalent, of course, to
$U \text{ is non-principal }\wedge(\forall z\lhd R((z\supsetneq U)\implies z\text{ is principal}))$

Also, you are probably assuming $R$ has identity, so it would never be the case that $R=z$, right? I think perhaps you are being confused by a quirk of the definition often given for maximal ideals.
By definition, a maximal ideal is
$M \text{ is a proper ideal }\wedge(\forall z\lhd R((z\text{ is a proper ideal })\wedge(z\supseteq M)\implies z=M))$
which can be restated as 
$M \text{ is a proper ideal }\wedge(\forall z\lhd R((z\supseteq M)\implies (z=M)\vee (z=R)))$
