\section{Introduction}\label{sec:01-introduction} The atomic ionization problem asks for the maximum number $\AtomCount{Z}$ of electrons that a nucleus of charge $Z$ can bind~\cite{nam2020,simon2000}. Experimentally, no isolated atom is known to bind more than one additional electron in a stable state~\cite{andersen2004}. The ionization conjecture asks for a bound $\AtomCount{Z}\leq Z+C$ with $C$ independent of $Z$. Lieb's bound $\AtomCount{Z}<2Z+1$~\cite{lieb1984} is sharp for hydrogen~\cite{bethe1929,hylleraas1930}, but permits four electrons for helium and six for lithium. Later work established asymptotic neutrality, $\AtomCount{Z}/Z\to1$ as $Z\to\infty$~\cite{lieb1988,fefferman1990,seco1990}, and improved explicit general bounds~\cite{nam2012,hundertmark2025}, without reducing these helium and lithium limits. Shelley and Tartakovsky recently excluded the helium dianion, lowering the helium upper bound to three electrons~\cite{shelley2026}. We prove that helium binds no additional electron in its ground state and lithium binds exactly one. We consider nonrelativistic electrons of spin $1/2$ and an infinitely heavy point nucleus, in Hartree atomic units. For an integer $N\geq1$ and $Z>0$, let $x_i\in\mathbb R^3$ be the position of electron $i$ relative to the nucleus. On the fermionic space $\FermionSpace{N}=\bigwedge^N L^2(\mathbb R^3;\mathbb C^2)$, set \[ \AtomH{N}{Z}=-\frac12\sum_{i=1}^N\Delta_{x_i} -Z\sum_{i=1}^N\frac1{|x_i|} +\sum_{i