Lang has been known to cloak theorems in the form of exercises, and this is an instance of this. What you are asking about is an elementary but important result in quadratic form theory, called Springer's Theorem.
(Beware: there is at least one other important result in this area called "Springer's Theorem". The other Springer's Theorem is Theorem 7 in these notes: it's a bit more technical.)
A quadratic form $q(x_1,\ldots,x_n)$ over a field $K$ is called anisotropic if for any vector $v \in K^n$ other than the zero vector, $q(v) \neq 0$. Then the contrapositive, hence equivalent, form of Lang's exercise is:
Theorem (Springer): Let $q$ be an anisotropic quadratic form over $K$, and let $L/K$ be an odd degree field extension. Then viewed as a quadratic form over $L$, $q$ remains anisotropic.
A proof of Springer's Theorem appears (e.g.) in $\S 3$ of these notes on quadratic forms. Despite the fact that this is almost 40 pages into a systematic study of the algebraic theory of quadratic forms, the proof should be self-contained provided the terminology is explained, which I hope I have done above.
Remark: A purely field theoretic result which can be viewed as a special case of Springer's Theorem is: if $K$ is a formally real field -- i.e., $-1$ is not a sum of squares -- and $L/K$ is a finite extension of odd degre, then $L$ is also formally real. (Indeed, a field is formally real iff for all $n$, the quadratic form $x_1^2 + \ldots + x_n^2$ is anisotropic.) This result appears in $\S 15.3$ of my field theory notes, and indeed the proof is essentially the same as the proof of Springer's Theorem.
Multivariate polynomials over an infinite field can have infinitely many roots, as pointed out by other users.
As for the univariate case, the answer is yes: if $f$ is a (univariate) polynomial over a field $K$ and $a\in K$ is a root, then we can use the division algorithm to show that $f(x)=(x-a)q(x)$ for some polynomial $q$ over $K$.
Note that this isn't the fundamental theorem of algebra, which says that every complex (univariate) polynomial of degree $n$ has exactly $n$ roots, counting multiplicity, in the complex numbers.
Best Answer
Let $F$ be an infinite field, and $f(x, y) \in F[x, y]$ a nonzero polynomial.
Regard $f$ as a polynomial $g(y) = f(x, y) \in (F(x))[y]$. This is a polynomial in $y$, with coefficients in the infinite field $F(x)$. Since it has a finite number of distinct roots, there is a $b \in F$ such that $0 \ne g(b) = f(x, b) \in F[x]$. Now apply the result for the univariate case.