No, by definition of "polynomial", a polynomial must have a largest, finite, power.
You can, however, have a "power series" that is of the form \(\displaystyle a_0+ a_1x+ a_2x^2+\cdot\cdot\cdot+ abx^n+ \cdot\cdot\cdot= \sum_{n=0}^\infty a_nx^n\) or \(\displaystyle a_0+ a_1(x- x_0)+ a_2(x- x_0)^2+\cdot\cdot\cdot+ ab(x- x_0)^n+ \cdot\cdot\cdot= \sum_{n=0}^\infty a_n(x- x_0)^n\) for some fixed number \(\displaystyle x_0\). The "McLaurin series" and 'Taylor series" are methods of forming such a power series corresponding to a given infinitely differentiable function.
Technically that is not a "polynomial". You would also need to answer questions of "convergence" and "divergence". For a power series there exist some "radius of convergence", R such that the series converges uniformly for \(\displaystyle |x- x_0|< R\). For example, the power series, \(\displaystyle \sum_{n=0}^\infty nx^n= x+ 2x^2+ 3x^3+ \cdot\cdot\cdot+ nx^n+ \cdot\cdot\cdot\) has "radius of convergence", R= 0 because it converges only for x= 0. But \(\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}\)\(\displaystyle = 1+ x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot\) has "radius of convergence" infinity because it converges for all x (and, in fact, is equal to \(\displaystyle e^x\) for all x).
You also cannot have a polynomial, or any function from R to R, that allows x to be "infinity" or takes on the value "infinity" because "infinity" is not a member of R, the set of real numbers. You can "extend" the real numbers to include "infinite points" but then you have trouble with basic arithmetic!
