A polynomial is completely determined by any part of it
If $p$ and $q$ are polynomials agreeing on infinitely many points, then $p-q$ is a polynomial that’s 0 on infinitely many points.
But if a polynomial $f$ of degree $n$ is $0$ on more than $n$ points, then it’s zero everywhere. (If it has zeroes $a_1, \ldots a_n$, then by repeated division it’s of the form $c(x-a_1)\cdots(x-a_n)$; if it’s zero at some other point as well, then we get $c=0$.)
So a polynomial that’s 0 on infinitely many points is 0 everywhere. So, going back to the beginning, if $p$ and $q$ are polynomials agreeing on infinitely many points, then $p-q$ is zero everywhere, i.e. $p=q$.
Polynomials are analytic functions. If two analytic functions agree on a set having a limit point, they must be equal by the Identity Theorem.
“There is one and only one polynomial” means two things:
1) There is at most one polynomial.
2) There is at least one polynomial.
Only the first affirmation is true.
1) There is at most one polynomial:
Proof by contradiction.
Assume $P$ and $Q$ are two different polynomials passing trough $(x_i,y_i)$, $i\in\mathbb N$. Let $n$ be the maximum of their degrees.
There is a unique polynomial of degree at most $n$ through $(x_i,y_i)$ for $i=1,...,n+1$. But we already know that $P$ and $Q$ do. So $P=Q$.
2) There may be no polynomial.
Example: let us consider the points $(n,e^n)$, $n\in\mathbb N$. Suppose $P$ is a polynomial passing through them.
Notice that $f(x)=e^x$ passes through them as well.
Hence if $$\lim_{x\to+\infty}\frac{e^x}{P(x)}$$ exists, it must be $1$, since it is $1$ when $x\in\mathbb N$. But (as it is easily proved using De L'Hospital), that limit exists and is $\pm\infty$. Contradiction. Hence, there is no such polynomial.
Conclusion: It is false that there exists a polynomial that passes through any infinite set of points, but if you know that the function is a polynomial beforehand, then it is uniquely determined.