Is boolean validity a harder problem than satisfiability?
However to check unsatisfiability I would need to check all possible assignments, surely this makes the problem EXPTIME?


Well, $EXPTIME$ but not necessarily $EXPTIME$-complete.

Congratulations, you've found the (possible) difference between $NP$ and $Co-NP$.

The first thing to realize is that there's a duality here. Satisfiability requires only a single example to prove a YES, but (as far as we know) needs to search a possibly exponential space for a NO. Unsat is the exact dual of this: you need to search a possibly exponential set to prove a YES, but a single counter-example will prove a NO. So neither is a harder problem, it's just that one is easy for YES, and the other is easy for NO.

The second thing to realize is that every NP-complete problem has a dual like this, and that's exactly why we study $Co-NP$ complete problems. For Hamiltonian Path, one path proves it, and a full search is needed to disprove. The converse is true to for NOHAMPATH, it's complement problem. Every NP-problem has a dual by switching the "yes" and "no" cases.

Nobody knows if $NP = EXPTIME$, though it's pretty unlikely. But it seems like just having to do an exponential search isn't enough to characterize all exponential algorithms.