I think these are all pretty standard, but context matters. I don't think there's a clear answer to your "particular" question, and I'll try to explain why.
Usually our usage of these terms comes from algorithm complexity. In terms of time complexity of algorithms, there are two definitions of exponential:
- EXPTIME: $2^{n^{O(1)}}$. That is, $2$ raised to a polynomial in $n$.
- E: $2^{O(n)}$. That is, $2$ raised to a function of $n$ that is at most linear.
Of course $2^{\sqrt{n}}$ fits in both classes. So the question is easily yes if we are just thinking about upper bounds: Naturally $2^{\sqrt{n}} < 2^n$ should be consider to lie in the class of exponential running times.
What I'm not so clear on is lower bounds. Specifically, I am not sure what a generally accepted interpretation of the following statement would be:
Deciding language $A$ requires exponential time.
(For example: If we show that $A$ requires time $2^{\sqrt{n}}$, should we say that $A$ requires exponential time?)
If we are thinking in the style of EXPTIME, I think we would say "yes": interpret the statement as saying that $A$ requires $2^{n^c}$ time for some $c > 0$. But thinking in the style of E, we would say no. For instance, the Exponential Time Hypothesis states informally that 3SAT requires exponential time, and formally it says that 3SAT requires $2^{cn}$ time for some $c$ (note the crucial difference!).
Because of this, I am not sure if there is a commonly accepted interpretation of the above statement. Ideally we would state more precisely what we mean when using the term, unless it's clear from context. Perhaps others can mention if they think one interpretation or the other is more "standard".
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In summary: if $f(n) \leq 2^{\sqrt{n}}$, then $f$ is definitely at most exponential. But if $f(n) \geq 2^{\sqrt{n}}$, it is not clear to me whether we would say $f$ is at least exponential.