= Solution
Because $F:\mathbf{Set}\to\mathbf{Set}$ preserves finite limits and colimits, $F(1)$ is a singleton and $F(\varnothing)=\varnothing$. For $x\in X$, let $\bar x:1\to X$ select $x$ and define
$$
\alpha_X(x)=F(\bar x)(*),
$$
where $*$ is the unique element of $F(1)$. This is natural in $X$. Any <natural transformation> $1_{\mathbf{Set}}\to F$ has the unique possible component at $1$, and naturality along every $\bar x$ forces the displayed value, so $\alpha$ is unique.
If $x\ne y$, the <equalizer> of $\bar x,\bar y:1\rightrightarrows X$ is empty. Since $F$ preserves this equalizer, $F(\bar x)(*)\ne F(\bar y)(*)$. Hence every $\alpha_X$ is injective, so $\alpha$ is pointwise monic.
Finite-colimit preservation makes $\alpha_n:n\to F(n)$ bijective for every finite set. Use the stated characterization of $\mathbb N$ by the coproduct diagram
$$
1\xrightarrow{0}\mathbb N\xleftarrow{s}\mathbb N
$$
and the coequalizer of $s,1_{\mathbb N}:\mathbb N\rightrightarrows\mathbb N$. Applying $F$ preserves both diagrams. Naturality and uniqueness in this characterization identify $\alpha_{\mathbb N}:\mathbb N\to F(\mathbb N)$ as an isomorphism.
For a countable family $(A_n)$, let $p:\coprod_nA_n\to\mathbb N$ record the summand. Each square
$$
\begin{array}{ccc}
A_n&\longrightarrow&\coprod_kA_k\\
\downarrow&&\downarrow p\\
1&\xrightarrow{n}&\mathbb N
\end{array}
$$
is a <pullback in a category>[pullback]. Applying $F$ and using $F(1)\cong1$ and $F(\mathbb N)\cong\mathbb N$ shows that $F(A_n)$ is exactly the fiber of $F(p)$ over $n$. Those fibers partition $F(\coprod_nA_n)$, so the canonical map
$$
\coprod_nF(A_n)\longrightarrow F\left(\coprod_nA_n\right)
$$
is bijective. Thus $F$ preserves <countable coproduct>[countable coproducts].
Now choose $x\in F(A)\setminus\alpha_A(A)$ and define
$$
U=\{A'\subseteq A:x\in\operatorname{im}(F(A')\to F(A))\}.
$$
It is upward closed. Since $A=A'\sqcup(A\setminus A')$ and $F$ preserves binary coproducts, $x$ lies in exactly one of the two summands, so exactly one of $A'$ and its complement lies in $U$. Pullback preservation gives closure under finite intersections.
For countable completeness, take $A_n\in U$ and put $B=\bigcap_nA_n$. If $B\notin U$, then $A\setminus B\in U$. Partition $A$ into $B$ and the sets
$$
C_n=(A\setminus A_n)\setminus\bigcup_{k<n}(A\setminus A_k),
$$
which record the first failed membership. Since $F$ preserves countable coproducts, exactly one cell of this partition lies in $U$. It cannot be $B$, so some $C_n\in U$. But $C_n\subseteq A\setminus A_n$, forcing $A\setminus A_n\in U$, contrary to $A_n\in U$. Hence $B\in U$.
Finally no finite $K$ belongs to $U$. Indeed, $F(K)\cong K$ through $\alpha_K$, so if $x$ came from $F(K)$ then naturality would put $x$ in the image of $\alpha_A$. Thus $U$ is a <countably complete ultrafilter> and is <nonprincipal ultrafilter>[nonprincipal].
Conversely, let $U$ be such an ultrafilter on $A$ and define the <ultrapower endofunctor of sets>
$$
F(B)=B^A/{\sim_U}.
$$
It preserves the terminal object and products: the map
$$
[(f,g)]\longmapsto([f],[g])
$$
is bijective because $U$ is closed under finite intersections. It preserves equalizers because an equality holding for an equivalence class holds on a $U$-large set, and the representative can be changed off that set to land in the equalizer. Hence it preserves all finite limits.
For $f:A\to\coprod_nB_n$, countable completeness implies that one index fiber
$$
\{a:f(a)\in B_n\}
$$
belongs to $U$; otherwise the countable intersection of all complementary fibers would be empty and belong to $U$. Thus every class $[f]$ lies in one and only one $F(B_n)$, proving preservation of countable coproducts.
The class of the identity map $A\to A$ is not represented by a constant map, since every equality set $\{a:a=a_0\}$ is a singleton and is not in $U$. Therefore $\alpha_A:A\to F(A)$ is not surjective. Since $\alpha$ is the unique natural transformation from the identity functor to $F$, a natural isomorphism $1_{\mathbf{Set}}\cong F$ would have to equal $\alpha$, which is impossible.
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