If a formal arithmetical object (for example, a power series in $\mathbb{Z}[[T]]$, or a formal subscheme of an arithmetic scheme) admits a complex meromorphic uniformization by meromorphic functions of finite order living on a sufficiently large region (an example: in the context of elliptic curves and their associated objects, consider the Weierstrass $\wp$ function, which is meromorphic of order $2$ on all of $\mathbb{C}$), then the formal object is algebraic.
Ultimately, such statements boil down to the product formula. Very roughly speaking, the complex uniformization yields a growth condition on "the coefficients" of a formal object, which collides with the integrality assumption to yield that almost all "coefficients" vanish, thereby yielding the desired algebraic relation.
Theorem. (J.-B. Bost) Let $G$ be an algebraic group over $\mathbb{Q}$. Consider a Lie subalgebra $\mathfrak{h} \subset \mathrm{Lie}(G)$ whose reductions $\mathfrak{h}\mod{p}$ at almost all primes $p$ are stable under $p$-th powers. Then there is an algebraic subgroup $H$ of $G$ such that $\mathfrak{h} = \mathrm{Lie}(H)$.
The statement of the opening paragraph is an easy consequence. Let $E,E'$ be elliptic curves over $\mathbb{Q}$ having $|E(\mathbb{F}_p)| = |E'(\mathbb{F}_p)|$ for almost all $p$. Apply Bost's theorem with $G:= E \times E'$ and $\mathfrak{h}$ an arbitrary one-dimensional horizontal $\mathbb{Q}$-vector subspace of $\mathrm{Lie}(G) = \mathrm{Lie}(E) \oplus \mathrm{Lie}(E')$. The $p$-th power operation on $\mathrm{Lie}(E)$$\mathrm{Lie}(E_p)$ (resp. $\mathrm{Lie}(E')$$\mathrm{Lie}(E'_p)$) is induced by Frobenius, and is equal mod (mod $p$) to multiplication by $1 -|E(\mathbb{F}_p)|$ (resp., by $1-|E'(\mathbb{F}_p)|$). If $|E(\mathbb{F}_p)| = |E'(\mathbb{F}_p)|$ then $\mathfrak{h}$$\mathfrak{h}\mod{p}$ is stable under the $p$-th power operation, and the conclusion follows from Bost's theorem.