Do finite algebraically closed fields exist?

No, there do not exist any finite algebraically closed fields. For suppose $K$ is a finite field; then the polynomial $$f(x)=1+\prod_{\alpha\in K}(x-\alpha)\in K[x]$$ cannot have any roots in $K$ (because $f(\alpha)=1$ for any $\alpha\in K$), so $K$ cannot be algebraically closed.

Note that for $K=\mathbb{F}_{p^n}$, the polynomial is $$f(x)=1+\prod_{\alpha\in K}(x-\alpha)=1+(x^{p^n}-x).$$

Hint $\rm\:F[x]\:$ has infinitely many primes for every field $\rm\,F\,$ -- by mimicking Euclid's proof for integers. In particular, if $\rm\,F\,$ is algebraically closed, there are infinitely many nonassociate primes $\rm\ x - a_i\:$ therefore there are infinitely many elements $\rm\:a_i\in F\:.\:$

Remark $\ $ This explains the genesis of the polynomial employed in Zev's answer.

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.