$\DeclareMathOperator\Spec{Spec}$Actually, your suggested categorical characterization of spectra of fields does work.
Edit: (I had written something incorrect here)
By Martin's comment below, we just have to show that maps from affines into $\Spec(k)$ are epis in the full category. But if we had two maps $f,g: \Spec(k) \rightarrow Y$ which agreed on some affine mapping into $\Spec(k)$, then first of all $f$ and $g$ would have to be the same topological map. Then both would land inside some affine $\Spec(R)\subset Y$, and now we're
reduced to the affine situation where we know it holds.
Conversely, suppose $X$ is not the spectrum of a field. If every point is dense, $X$ is affine
and we are done by what we know about the affine subcategory. Otherwise, we can find an open subscheme $U\subsetneq X$. Then the inclusion of $U$ into $X$ is not an epi as is witnessed by the two inclusions
$$X \rightrightarrows X\sqcup_U X,$$
where the last object is $X$ glued to itself along $U$.
Theorem: Every epi is surjective.
Proof.
Let $h : A \to B$ be an epimorphism. We define maps $f, g : B \to \mathcal{P}(B)$ by
\begin{align*}
f(b) &= \{b\} \cap \mathrm{im}(h)\\
g(b) &= \{b\}
\end{align*}
where we recall that $\mathrm{im}(h) = \{b \in B \mid \exists a \in A \,.\, h(a) = b\}$.
For every $a \in A$ we have $f(h(a)) = \{h(a)\} = g(h(a))$, therefore $f = g$ as $h$ is epi. Now, for every $y \in B$ we have $\{y\} = g(y) = f(y) = \{y\} \cap \mathrm{im}(h)$, therefore $y \in \mathrm{im}(h)$. QED.
Supplemental 2022-01-09: Here is an improved version which uses a smaller codomain. We write $\Omega$ for the subobject classifier (the set of truth values).
Proof.
Let $h : A \to B$ be an epimorphism and $b \in B$. We define maps $f, g : B \to \Omega$ by
\begin{align*}
f(b') &{}\mathbin{{:}{=}} (b = b' \land \exists a \in A . h(a) = b')\\
g(b') &{}\mathbin{{:}{=}} (b = b').
\end{align*}
For every $a \in A$ we have
\begin{align*}
f(h(a)) &\Leftrightarrow (b = h(a) \land \exists a' \in A . h(a') = h(a)) \\
&\Leftrightarrow (b = h(a)) \\
&\Leftrightarrow g(h(a)),
\end{align*}
therefore $f = g$ as $h$ is epi. Now
\begin{align*}
\top &\Leftrightarrow b = b \\
&\Leftrightarrow g(b) = f(b) \\
&\Leftrightarrow b = b \land \exists a \in A . h(a) = b \\
&\Leftrightarrow \exists a \in A . h(a) = b. \quad \Box
\end{align*}
Best Answer
Update: the following exchange appeared on the categories mailing list several years ago: http://article.gmane.org/gmane.science.mathematics.categories/3094. Walter Tholen's response strongly suggests that the answer to Martin's question is that the converse does not hold, although I don't have access to the four-author article he cites as reference. It's probably worth a look though, and if I learn anything more I'll post another update.
Second update: My surmise was correct. Walter Tholen kindly emailed to me the relevant two pages (pp. 88-89) of the four-author paper
where the following example is given on page 89: in the category of semigroups with zero such that all 4-fold products are zero, all epimorphisms are surjective but not all monos are regular [a specific nonregular mono is described]. (I can forward this email if you write to me at topological dot musings at gmail dot com.)
Assuming amalgamated products exist (as they do in categories of algebras of a Lawvere theory), a mono $i: A \rightarrowtail B$ is regular if it is the equalizer of the pair of canonical maps from $B$ to the amalgamated product $B *_A B$ (i.e., the coprojections of the pushout of $i$ with itself, aka the cokernel pair of $i$). The equalizer of the cokernel pair defines a closure operator on the lattice of subalgebras $\mathsf{Sub}(B)$, called the dominion operator $\mathsf{Dom}_B$. So to prove a subalgebra is not regular is to show that it is not $\mathsf{Dom}$-closed. The key technical result needed to prove the claim above is Isbell's Zig-Zag theorem (given in his paper Epimorphisms and Dominions in the 1965 La Jolla conference proceedings on categorical algebra), as recalled in Peter M. Higgins. "A short proof of Isbell's Zigzag Theorem." Pacific J. Math. 144(1):47–50 (1990), which gives a precise and useful criterion for an element to belong to the dominion (i.e., the $\mathsf{Dom}$-closure) of a subalgebra.
Hope this helps. I am voting up your question, Martin, since it's rather nontrivial!