Zero divisors of $\mathbb{Z}_{7}[x] / (x^4+x^3-3)$ and inverse element of $\overline{x+1}$

I answer assuming $\mathbb Z_7$ is the finite field of $7$ elements. The polynomial $f=x^4+x^3-3$ factors as $(x + 4) (x + 5) (x^2 + 6x + 3)$ to irreducibles. So by the Chinese Remainder Theorem we have $$\mathbb Z_7[x]/(f) \cong \mathbb Z_7[x]/(x+4)\times \mathbb Z_7[x]/(x+5) \times \mathbb Z_7[x]/(x^2+6x+3)\cong \mathbb Z_7\times \mathbb Z_7\times \mathbb F_{49},$$ where $\mathbb F_{49}$ is the finite field with $49$ elements. The zero divisors are the preimages of $(x,y,z)$ with at least one of $x,y,z$ being zero.

To have explicit description, you need to write down the CRT formula.

Edit: I thought the question was to find the zero divisors. To compute how many there are, we note that in a finite ring a nonzero element $x$ is either invertible or a zero divisor (since the multiplication-by-$x$ map is either injective and hence from finiteness also surjective, hence invertible, or it has a non-trivial kernel and so the element is zero divisor). Now an element is invertible if and only if it is invertible in each of the components in the factorization above. So we have $6*6*48= 1728$ invertibles and $7^4-1728-1= 672$ nonzero zero-divisors.