As has been pointed out, with the way you have worded things, there is no bijection, since one of your collections of polynomials is infinite, while the other collection is empty.
Let us look at a generalization of the problem. Suppose you have an explicit listing of two types of polynomials with integer coefficients, Type $1$ and Type $2$ (in your case, Type $1$ is all polynomials with integer coefficients, Type $2$ is positive integer coefficients).
Now suppose that you want to exhibit a bijection between Type $1$ polynomials that satisfy a certain additional condition $A$, and Type $2$ polynomials that satisfy an additional condition $B$. This will not be possible if the cardinalities of the two collections are different. So suppose for example that there are infinitely many Type $1$ polynomials that satisfy condition $A$, and infinitely many Type $2$ polynomials that satisfy condition $B$.
We can list the Type $1$ polynomials that satisfy condition $A$ as $P_1$, $P_2$, $P_3$, and so on, by letting $P_1$ be the first polynomial in the listing of Type $1$ polynomials that satisfies $A$, letting $P_2$ be the second one, and so on.
Similarly, we can list the Type $2$ polynomials that satisfy condition $B$ as $Q_1$, $Q_2$, $Q_3$, and so on by using the same strategy.
Now we get an explicit bijection between Type $1$ polynomials that satisfy $A$ and Type $2$ polynomials that satisfy $B$ by mapping $P_k$ to $Q_k$ for all $k$.
Is this bijection really explicit? Suppose we have a computer program for listing Type $1$ polynomials, and a computer program that can recognize whether a polynomial satisfies condition $A$. Suppose also that we have similar programs for Type $2$ polynomials and for recognizing condition $B$. Then we can write an explicit computer program that will produce the bijection.
In general, one cannot expect that the bijection will be given by a simple short formula. But what I have outlined is more than an existence proof, it is, under appropriate conditions, a recipe.