Does any integral domain contain an irreducible element?

Hint $\ $ There are rings, not fields, closed under square roots (e.g. the ring of all algebraic integers). In such a ring there are no irreducibles since every element factors: $\rm\ a = \sqrt{a} \sqrt{a}.$

A concrete example: $ $ let $\rm\, D = \Bbb Q[x,x^{1/2},x^{1/4},x^{1/8},\ldots].\,$ Every element $\ne0$ is writable in the form $\rm\ d_i x^{i}\! + d_j x^j\! +\cdots\! + d_k x^k\! = x^i\, f\ \,$ for $\rm\,\ f\in D,\,\ d_i\in\Bbb Q,\,\ d_i\ne 0,\:$ and $\rm\:i < j < \ldots < k,\:$ e.g.

$$\rm a\, x^{1/8}\! + b\, x^{1/4}\! + c\, x^{1/2} =\, x^{1/8} (a + b\, x^{1/8}\! + c\, x^{3/8})\, =\, x^{1/8}\, f,\quad a\ne 0$$

All exponents of $\rm\,x\,$ in terms of $\rm\,f\,$ remain of form $\rm\,n/2^m\!,\,$ being differences of such. We can force all the cofactors $\rm\,f\,$ to be units by adjoining inverses of all such elements, as follows. Since $\rm\,f\,$ has "constant" term $\rm\:a\ne 0,\,\ f\,$ is not in the maximal ideal $\rm\, M = (x,x^{1/2},x^{1/4},x^{1/8},\ldots).$ Therefore, localizing at $\rm\,M,\,$ i.e. adjoining inverses of all elements $\rm\not\in M,\,$ turns all the $\rm\,f\,$ cofactors into units, yielding a domain $\rm\,\bar D,\,$ not a field $\rm(x^{-1}\!\not\in \bar D),\,$ where all elements have form $\rm\,x^k\, u,\,$ for some unit $\rm\,u.\,$ Since $\rm\ a = x^k = (x^{k/2})^2 = \sqrt{a}\sqrt{a},\ $ this domain has no irreducibles.

It is proven here that the ring of all algebraic integers has no irreducible elements.

Another class of examples is given by valuation rings whose value group is divisible.