Would a parser for this left-recursive grammar require infinite input?
Your grammar captures the empty language. To see this, you need to use the definition of a language described by a context-free grammar.
A context-free grammar is a tuple $(T,N,S,P)$, where $T$ is the set of terminals (a non-empty finite set), $N$ is the set of nonterminals (a non-empty finite set disjoint from $T$), $S \in N$ is the starting symbol, and $P$ is a collection of productions $A \to \alpha$, where $A \in N$ and $\beta \in (N \cup T)^*$.
In your case, $T = \{a\}$, $N = \{\mathrm{Expr}\}$, $S = \mathrm{Expr}$ (you are not stating this explicitly, but this is the only choice), and $P$ consists of the single rule $\mathrm{Expr} \to \mathrm{Expr} \, a$.
We say that $\beta$ produces $\gamma$ in one step, in symbols $\beta \Rightarrow \gamma$, if there exists a production $A \to \alpha \in P$ and two strings $\lambda,\mu \in (N \cup T)^*$ such that $\beta = \lambda A \mu$ and $\gamma = \lambda \alpha \mu$.
We say that $\beta$ produces $\gamma$, in symbols $\beta \Rightarrow^* \gamma$, if either $\gamma = \beta$, or $\beta \Rightarrow \delta \Rightarrow^* \gamma$ for some $\delta \in (N \cup T)^*$. This is the reflexive-transitive closure of $\Rightarrow$.
Finally, the language produced by the grammar is $\{ w \in T^* : S \Rightarrow^* w \}$.
In your case, a simple induction shows that $\mathrm{Expr} \Rightarrow^* \alpha$ iff $\alpha = \mathrm{Expr} \, a^n$ for some $n \in \mathbb{N}$. Since $\mathrm{Expr} \, a^n \notin T^*$, we conclude that the language produced by your grammar is the empty language.