How can I check whether the group $[16,13]$ in GAP with $3$ generators can be generated by $2$ elements?

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The group $[16,13]$ in GAP has structure $(C4\times C2):C2$ and is generated by the permutations $(1234)(5678)$ , $(15)(26)(37)(48)$ and $(57)(68)$ . The group $[16,3]$ in contrast with the same structure is a $2$-generated group : $(12)(34)$ and $(23)(5678)$ is a possible set of generators.

How can I prove that the group $[16,13]$ cannot be generated by $2$ elements ?

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In gap there is a MinimalGeneratingSet(G) command which pulls minimal generating set, in this case it says it is a three element set. There is some more information on this page. The documentation says it is really only efficient for nilpotent, solvable groups, and some special cases (if you have all the correct packages).

In this particular case you could also use gap to compute the abelianization (or by hand possibly), which is rank three (and order 8 group), so the full group could not be generate by less than three elements. A general strategy I can think of is that you can compute quotients, and figure out the minimal generating set of the quotients (which in principal would be easier) and get bound on the minimal number of generators required.