Suppose such polynomial exists, with degreen n > 0:
P(x) = anxn + an-1xn-1 + ... + a1x + a0
with P(k) prime for all k = 0, 1, 2, ...
Now an = P(0), so it's a prime; let it be p.
Now observer that P(ip) is a multiple of p for every i = 0, 1, 2, ... since p divides each term of the polynomial.
But these numbers P(ip) for i = 0, 1, 2, ... are also primes. The only way that they can be prime and
a multipe of p is for them to equal p. Thus P(ip) = p for i = 0, 1, 2...
Now consider the polynomial Q(x) = P(px) - p.
Q(x) is also of degree n, and has infinitely many roots (it has a root at every non-negative integer x = 0, 1, 2, ...).
But this is a contradiction: a polynomial of degreen n has at most n roots.
Hence there can't be a polynomial P(x) for which P(k) is prime for all k = 0, 1, 2, ...