I am not exactly a group theorist, so this may be well-known.
Let $G$ be a finitely generated group such that the cardinality of minimal generating sets of $G$ is bounded above. Does it follow that $G$ is finite?
This is true if $G$ is abelian, but I have no idea about the general case.
Just to expand my comment, a Tarski Monster is an infinite group in which, for some fixed prime $p$, all proper nontrivial subgroups have order $p$. It was proved by Olshanskii in 1979 that they exist for all primes $p>10^{75}$.
A set of $p+1$ distinct elements of such a group cannot all lie in the same subgroup of order $p$, so they must generate the whole group, and so we have an upper bound of $p+1$ on the size of minimal generating sets.
It would interesting to know whether there are any less exotic examples.
There is a simpler construction indeed, namely the Grigorchuk group $\Bbb G$. Though still not very easy, it's definitely "less exotic". For a readable introduction, see our mini-survey, or even better Ch. VIII in de la Harpe's monograph.
Specifically, on your question, it is known that every generating set with 4 elements is redundant. This follows immediately from Pervova's computation of the Frattini subgroup $\Phi(\Bbb G)$ given in this paper. She proved that all maximal subgroups of $\Bbb G$ have index 2 in $\Bbb G$. This easily implies
$$\Bbb G/\Phi(\Bbb G) = \Bbb G/[\Bbb G,\Bbb G]= \Bbb Z_2^3.$$
For more on the context of Pervova's work and other results on the Grigorchuk group, see this survey by Grigorchuk (esp. $\S 6$).
It is also worth noting that you shouldn't expect much easier examples. If even one minimal generating set has an element $g$ of infinite order, one can use powers of $g$ to make large minimal generating sets. Unsurprisingly, all elements in $\Bbb G$ have finite orders, which are in fact powers of 2.
This is a variant of Derek Holt's answer. Note that if $X$ is a finite group, and $S = \{x_{1},x_{2},\ldots,x_{r} \}$ is an irredundant generating set for $X$ ( that is, if any $x_{i}$ is omitted, the remaining elements do not generate $X$), then $\langle x_{1} \rangle < \langle x_{1},x_{2} \rangle < \ldots \langle x_{1},\ldots x_{r-1} \rangle < X$ is a strictly increasing sequence of subgroups of $X.$ It follows that $r \leq \log_{2}(|X|),$ in fact $r \leq \log_{p}(|X|),$ where $p$ is the smallest prime divisor of $|X|.$ Hence if $G$ is an infinite group, all of whose proper subgroups are finite, of order at most $n,$ then any irredundant generating set for $G$ has at most $1 + \log_{2}(n)$ elements.For if the chosen generating set has $s$ elements, then any $s-1$ elements irredundantly generate a proper subgroup, say $H$, of $G$, and $H$ has order at most $n.$ This bound can obviously be improved with finer knowledge of the possible prime divisors of orders of proper subgroups of $G.$
