Given a generating function (ordinary, exponential, or otherwise) such as $$ G(a_n;\, x) := \sum_{n=0}^\infty a_nx^n $$ where one or more $a_n = 0$, is there any way to prove that $G$ does or does not have an infinite number of zeros (i.e., there is no maximum $m$ such that $a_m=0$)?
For example, let's say I have a $G$ which yields $$ G(a_n;\, x) = a_2x^2 + a_3x^3 + a_5x^5 + a_6x^6 + a_7x^7 + a_8x^8 + a_{10}x^{10} + \dotsb $$ where the gaps (“zeros”) are exactly when $n$ is an integer square — leaving aside for the moment the obvious fact that this clearly isn't the best way to generate the set of squares in a power series, can the definition of $G$ be used to prove that the set of integer squares is infinite?
In the broadest sense, no, there cannot be such a procedure. Let's assume it were decidable whether there is a maximal $m$ such that $a_m = 0$. Then we could take a generic enumeration of all Turing Machines and define a generating function $A(z)$ via $a_m = 0$ if the $m$th machine $M_m $ halts on input $M_m$, and $1$ otherwise.
Now if we could decide whether there can be a maximal such $a_m$, we could also solve the Halting Problem.
Nevertheless, there will be many cases when you can decide that there is a maximal zero coefficient. Mostly that will happen by a simple recurrence relation on the coefficients (think of the recurrence relation for the Fibonacci or Catalan numbers), but in its broadest sense you are asking whether it's decidable if a general number sequence has a finite number of zeros, and that problem is undecidable.