If I were given the points
(1,3) (2,5) (3,7) and assumed this pattern continued forever, I know that it is linear as there is a constant the y value for an increase of one for the x value.
If I were given the points
(-2,4) (-1,1) (0,0) (1,1) (2,4)
I know that this is quadratic because the differences of the differences of the y values is 2 while the x value increase by one each time.
With only points, I can I prove a graph is exponential?
This is a little difficult to answer, because if I gave you three points $(1.7,3.9), (3.1, 6.1)$, and $(5.123, 8.99)$, these three points do actually all lie on the same parabola (and are thus quadratic). Nevertheless, I know what you are asking, and I'll try to answer clearly. To skip to the direct answer, you can skip this part.
In your second example, I notice that you picked a very nice quadratic function, namely $f(x) = x^2$. I would like to pick an equally nice exponential function, say, $g(x) = 2^x$. Let's also choose a few... convenient $x$-coordinates, $x=0,1,2,3$. Plugging in, we obtain a list of points similar to those in your second example, $(0,1), (1,2), (2,4), (3,8)$. Your question seems to be asking, "Given a set of points, can I reverse engineer this process to obtain an exponential function?"
And in some cases, the answer is yes! (As in above)
The trick here is your choice of $x$-coordinates. Specifically, in both examples you chose evenly-spaced $x$-coordinates. I know you're asking about points which were given to you, but the idea is to search for a pattern, and to do that we may need some handouts for now. Looking now at your first example, you noticed that every time $x$ increased by $1$, $y$ increased by $2$. The name for this pattern is called "linear"!
If you look at my points above, you notice that every time $x$ increases by $1$, $y$ is $multiplied$ by some value, namely $2$. Then I can write any $y$ value as its corresponding power of $2$, so for example, $64 = 2^6$, and so $(6,64)$ is part of this pattern.
As an example, say I was given $(3,125), (4,625), (5,3125)$. I notice that as $x$ increases by $1$, $y$ scales by 5, so this function should behave like $5^x$.
One last thing to mention is that just like with linear patterns and quadratic patterns, there are variants of the "easy functions" such as $y=a^x$ for some $a$. For example, the lists given by:
$(0,2), (1,3), (2,5), (3,9)$ ~ $2^x+1$
$(1,10), (2,50), (3,250), (4,1250)$ ~ $2*5^x$
$(0,9), (1,27), (2,81), (3,243)$ ~ $3^{x+2}$
are all exponential. This can cause an issue in trying to identify which exponential function fits these points (as in the first example, the scaling of the $y$ values isn't observed directly from the points themselves).
Apologies for the long response, but here is the direct answer. you mentioned that the change in $y$ values being constant implies that the function is linear, and the change in $y$ values being linear means that the function is quadratic.
The desired property for exponential functions is that the change in $y$ values is... still exponential! Computing these changes for my three examples, we see
$(0,2), (1,3), (2,5), (3,9) \rightarrow 1,2,4,...$
$(1,10), (2,50), (3,250), (4,1250)\rightarrow 40, 200, 1000,... $ ~ $40*5^x$.
$(0,9), (1,27), (2,81), (3,243) \rightarrow 18, 54, 162... $ ~ $18*3^x$.
With the second two examples, this process is unnecessary. But with the first, this process helps us clearly recognize the underlying exponential nature of these points. Again I apologize for the long post, but I hope this discussion aids in unveiling the patterns underlying exponential functions.