Since $\widehat h-h$ is bounded, a bound on $\widehat h(P)$ bounds the <projective height> of $x(P)$. The <Northcott theorem> gives only finitely many possible rational $x$-coordinates, and each has at most two points above it. Hence
$$
\boxed{N(B)<\infty.}
$$

By the <Mordell-Weil theorem>, $E(\mathbb Q)\cong\mathbb Z^r\oplus E(\mathbb Q)_{\mathrm{tors}}$. The canonical height is a positive-definite <quadratic form> on the lattice $\mathbb Z^r$, so <canonical-height lattice-point growth> gives
$$
N(B)=O(B^{r/2}).
$$
If $r=0$, this count is bounded; if $r=1$, it is $O(\sqrt B)$. Consequently
$$
\boxed{\frac{N(B)}{\sqrt B}\longrightarrow\infty\quad\Longrightarrow\quad\operatorname{rank}E(\mathbb Q)\geq2.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 125 / 5 / c / Solution

If an elliptic curve over $\mathbb Q$ has <Mordell-Weil group> of rank $r$, then its <canonical height of an elliptic curve> is a positive-definite <quadratic form> on the free lattice. Consequently
$$
\#\{P:\widehat h(P)\leq B\}=O(B^{r/2}).
$$

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 Incoming links: Canonical-height lattice-point growth