Grammar with fewest variables
Let $L \neq \emptyset$ be a language over some alphabet $\Sigma$, let $|$ be a symbol not in $\Sigma$, and consider the language $|L = \{ |w : w \in L \}$. Suppose that $L$ could be generated using a context-free grammar with a single nonterminal $S$.

Since $L \neq \emptyset$, $S \Rightarrow^* |w$ for some word $w$. Therefore:


Every rule contains at most one $S$ on the right-hand side. Indeed, otherwise you could generate a word with more than one $|$.
Every rule containing exactly one $S$ must be of the form $S \to Sw$, where $w \in \Sigma^*$. Indeed, the rule cannot contain $|$, since otherwise $S$ would generate a word with more than one $|$; and the rule cannot contain anything to the left of $S$ since then $S$ would generate a word not beginning in $|$.
Every rule not containing any $S$ must be of the form $S \to |w$, where $w \in \Sigma^*$ (in fact, $w \in L$).


Let $A$ be the collection of words $w \in \Sigma^*$ for which there is a rule $S \to |w$, and let $B$ be the collection of words $w \in \Sigma^*$ for which there is a rule $S \to Sw$. Thus
$$ L = |AB^*. $$
Conversely, it is easy to check that every language of this form (with $A,B$ finite) has a context-free grammar with a single nonterminal. Considering also the empty language, we have shown:


  The language $|L$ can be generated using a context-free grammar with a single nonterminal iff $L = AB^*$ for some finite (possibly empty) $A,B$.


The collection of languages of the form $AB^*$ is closed under quotient from the left, which is the following operation: $\sigma^{-1}S = \{ w : \sigma w \in S \}$. Indeed, if $\epsilon \notin A$ then $\sigma^{-1}(AB^*) = (\sigma^{-1}A)B^*$, and if $\epsilon \in A$ then $\sigma^{-1}(AB^*) = (\sigma^{-1}A)B^* \cup (\sigma^{-1}B)B^*$. Therefore, Greibach's theorem shows that determining whether a context-free grammar generates a language of the form $AB^*$ is undecidable. 

Given a context-free grammar $G$, we easily construct a context-free grammar for $|L(G)$. The language of this context-free grammar can be generated by a single nonterminal iff $L(G)$ is of the form $AB^*$, and so the following problem is undecidable: Given a context-free grammar, determine whether its language can be generated by a context-free grammar with a single nonterminal.