As others have said, there cannot be any finite algebraically closed fields (and if there were, algebraic geometry would be a rather different subject than it is;-). In fact there cannot even be any finite field $K$ over which all *quadratic* polynomials have roots, by the following simple counting argument.

Let $q=|K|$, then there are $q$ monic degree $1$ polynomials $X-a$, and similarly $q^2$ monic degree $2$ polynomials $X^2+c_1X+c_0$ in $K[X]$. By commutativity there are only $\frac{q^2+q}2$ distinct products of two degree $1$ polynomials, which leaves $q^2-\frac{q^2+q}2=\binom{q}2$ irreducible monic quadratic polynomials. (Even without using unique factorization, one gets *at least* so many monic irreducible polynomials.)

There are in fact formulas in terms of $q$ for the number of (monic) irreducible polynomials over $K$ of any degree, obtained by the inclusion–exclusion principle.

These formulas show that, if the mythological field with $1$ element were to exist, it would be algebraically closed.

**Added:** It turns out that finding the number monic irreducible polynomials over $K$ of a given degree using only inclusion–exclusion (and nothing about finite fields) gets rather hairy. Rather, one can use the existence of finite field $K'$ of order $q^n$ to find the formula. All elements of $K'$ have a minimal polynomial of degree $d$ dividing $n$, since they are contained in a subfield of order $q^d$, and inversely all $d$ roots of such an irreducible polynomial are distinct and lie in $K'$. Then if $c_d(q)$ counts the irreducible polynomials of degree $d$ over $K$, one has $\sum_{d|n}dc_d(q)=q^n$. From this a Möbius inversion argument (which is a form of inclusion–exclusion) gives
$$
c_n(q)=\frac{\sum_{d|n}\mu(n/d)q^d }n
$$
where $\mu$ is the classical Möbius function.