Let $\webleft (X,x_{0}\webright )$, $\webleft (Y,y_{0}\webright )$, and $\webleft (Z,z_{0}\webright )$ be pointed sets and let $f\colon \webleft (X,x_{0}\webright )\to \webleft (Z,z_{0}\webright )$ and $g\colon \webleft (Y,y_{0}\webright )\to \webleft (Z,z_{0}\webright )$ be morphisms of pointed sets.
We claim that $X\times _{Z}Y$ is the categorical pullback of $\webleft (X,x_{0}\webright )$ and $\webleft (Y,y_{0}\webright )$ over $\webleft (Z,z_{0}\webright )$ with respect to $\webleft (f,g\webright )$ in $\mathsf{Sets}_{*}$. First we need to check that the relevant pullback diagram commutes, i.e. that we have
Indeed, given $\webleft (x,y\webright )\in X\times _{Z}Y$, we have
\begin{align*} \webleft [f\circ \text{pr}_{1}\webright ]\webleft (x,y\webright ) & = f\webleft (\text{pr}_{1}\webleft (x,y\webright )\webright )\\ & = f\webleft (x\webright )\\ & = g\webleft (y\webright )\\ & = g\webleft (\text{pr}_{2}\webleft (x,y\webright )\webright )\\ & = \webleft [g\circ \text{pr}_{2}\webright ]\webleft (x,y\webright ),\end{align*}
where $f\webleft (x\webright )=g\webleft (y\webright )$ since $\webleft (x,y\webright )\in X\times _{Z}Y$. Next, we prove that $X\times _{Z}Y$ satisfies the universal property of the pullback. Suppose we have a diagram of the form
in $\mathsf{Sets}_{*}$. Then there exists a unique morphism of pointed sets
\[ \phi \colon \webleft (P,*\webright )\to \webleft (X\times _{Z}Y,\webleft (x_{0},y_{0}\webright )\webright ) \]
making the diagram
commute, being uniquely determined by the conditions
\begin{align*} \text{pr}_{1}\circ \phi & = p_{1},\\ \text{pr}_{2}\circ \phi & = p_{2}\end{align*}
via
\[ \phi \webleft (x\webright )=\webleft (p_{1}\webleft (x\webright ),p_{2}\webleft (x\webright )\webright ) \]
for each $x\in P$, where we note that $\webleft (p_{1}\webleft (x\webright ),p_{2}\webleft (x\webright )\webright )\in X\times Y$ indeed lies in $X\times _{Z}Y$ by the condition
\[ f\circ p_{1}=g\circ p_{2}, \]
which gives
\[ f\webleft (p_{1}\webleft (x\webright )\webright )=g\webleft (p_{2}\webleft (x\webright )\webright ) \]
for each $x\in P$, so that $\webleft (p_{1}\webleft (x\webright ),p_{2}\webleft (x\webright )\webright )\in X\times _{Z}Y$. Lastly, we note that $\phi $ is indeed a morphism of pointed sets, as we have
\begin{align*} \phi \webleft (*\webright ) & = \webleft (p_{1}\webleft (*\webright ),p_{2}\webleft (*\webright )\webright )\\ & = \webleft (x_{0},y_{0}\webright ),\end{align*}
where we have used that $p_{1}$ and $p_{2}$ are morphisms of pointed sets.