Is EXPTIME "solvable" or "checkable" in exponential time?
A language $L$ belongs to $\mathsf{EXPTIME}$ if there is an algorithm deciding it which runs in time $O(2^{n^k})$ for some $k$.

A language $L$ belongs to $\mathsf{NEXPTIME}$ if there is an algorithm verifying it which runs in time $O(2^{n^k})$ for some $k$. In other words, a language $L$ belongs to $\mathsf{NEXPTIME}$ if there is a constant $k$ and a machine $M$ running in time $O(2^{n^k})$ such that $x \in L$ iff there exists $y$ of size at most $O(2^{|x|^k})$ such that $M(x,y) = 1$. Alternatively, a language $L$ belongs to $\mathsf{NEXPTIME}$ if it can be decided in nondeterministic time $O(2^{n^k})$ for some $k$.

You observe correctly that every problem in $\mathsf{NP}$ has an $\mathsf{EXPTIME}$ algorithm, that just goes over all possible witnesses. This shows that $\mathsf{NP} \subseteq \mathsf{EXPTIME}$. It is conjectured that the two classes are distinct. In contrast, it is believed that $\mathsf{EXPTIME} \neq \mathsf{NEXPTIME}$, and this is stronger than (i.e., implies) the more famous conjecture $\mathsf{P} \neq \mathsf{NP}$, through a padding argument.

Note that nondeterministic computation can still be time-bounded. For example, $\mathsf{NP}$ is the class of languages computed by nondeterministic polytime algorithms. The nondeterministic time hierarchy theorem shows that $\mathsf{NP} \neq \mathsf{NEXPTIME}$; similarly, the deterministic time hierarchy theorem shows that $\mathsf{P} \neq \mathsf{EXPTIME}$.