Is it right to call $2^{\sqrt{n}}$ "exponential"?
In general I agree with usul's summary: for upper bounds $2^{\sqrt{n}}$ is certainly at most exponential, and for lower bounds it's more context-dependent and less clear. Let me offer a couple more examples from different contexts, and provide the term "moderately exponential" for your consideration.


In the literature on graph isomorphism (which is typically dealing with upper bounds), $2^{\sqrt{n} poly(\log n)}$ (or even just $2^{n^c}$ with $c &lt; 1$) is often referred to as moderately exponential (though I understand this term has a different meaning the context of exact algorithms for $\mathsf{NP}$-complete problems; for the relation between the two, see the Appendix to Paolo Codenotti's Thesis). In some sense one can view this as very humble, but not belittling, terminology: it recognizes how far such algorithms are from polynomial (or even quasi-polynomial) while still acknowledging that it represents quite significant progress over the brute force $n!$ algorithm, or even over singly-exponential ($2^{\Theta(n)}$) algorithms.


Referring to a $2^{\sqrt{n}}$ lower bound as moderately exponential could serve a similar function: humble enough to differentiate it from $2^{\Omega(n)}$ lower bounds, but significant enough to distinguish it from merely super-polynomial. On the other hand:


$2^{n^{O(1/d)}}$ lower bounds for depth $d$ Boolean circuits (or the corresponding Frege classes in proof complexity) are often referred to as "exponential lower bounds" to distinguish them from "merely" super-polynomial or quasi-polynomial lower bounds. However, as getting a $2^{\Omega(n)}$ lower bound in these situations can be quite important, the latter may sometimes be referred to as "strong exponential lower bounds" or sometimes "truly exponential lower bounds" to distinguish it from the "merely" $2^{n^{c}}$ lower bounds (with $c &lt; 1$). 
In the context of depth 4 algebraic circuits, $2^{\tilde{\Omega}(\sqrt{n})}$ lower bounds have been referred to as exponential. Note that getting any stronger lower bounds would separate $\mathsf{VP}$ from $\mathsf{VNP}$.


(Even in the context of Graph Isomorphism, I could see someone, perhaps informally, referring to a $2^{\sqrt{n}}$ lower bound as "exponential", since it would be understood that there is a $2^{O(\sqrt{n \log n})}$ upper bound so essentially no stronger exponential lower bound could be hoped for...)