Here's an example: consider the $\infty$-category $Fun(\Delta^1,Perf(\mathbb F_p))$. It has two canonical generators $A= \mathbb F_p\to \mathbb F_p$ and $B=\mathbb F_p\to 0$; and I claim that $R= End(A\oplus \Sigma B)$ is an example of what you're looking for (derived endomorphisms - more generally, everything here is "derived").

Note that $\hom(A,B) = \mathbb F_p$, $\hom(B,A)=0$, $\hom(A,A) = \mathbb F_p, \hom(B,B)= \mathbb F_p$.

In particular, $End(A\oplus\Sigma B) = \hom(A,\Sigma B)\oplus \hom(A,A)\oplus \hom(\Sigma B,A)\oplus \hom(\Sigma B,\Sigma B)$.

Note that $\hom(\Sigma B,\Sigma B)\simeq \hom(B,B)$, $\hom(\Sigma B,A) = 0$, so really $End(A\oplus \Sigma B)= \hom(A,A)\oplus \hom(B,B)\oplus \Sigma\hom(A,B)\simeq \mathbb F_p\oplus\mathbb F_p\oplus\Sigma\mathbb F_p$.

In particular, $End(A\oplus\Sigma B)$ is connective and perfect over $\mathbb F_p$, so it remains to argue that it is smooth. But smoothness is Morita invariant, i.e. it only depends on the category of modules.

Now, $A,B$ are generators of $Fun(\Delta^1,Perf(\mathbb F_p))$, so $Perf(End(A\oplus\Sigma B))\simeq Fun(\Delta^1,Perf(\mathbb F_p))$ by the Schwede-Shipley theorem (I'm using implicitly here that $Ind(C^{\Delta^1}) = Ind(C)^{\Delta^1}$ for any stable $C$), so it suffices to argue that $Fun(\Delta^1,Perf(\mathbb F_p))$ is smooth.

Now, for an indexing diagram $C$, $Fun(C,Mod_R)^\omega$ is smooth over $R$ if and only if $R[map_C(-,-)]$ is compact in $Fun(C^{op}\times C,Mod_R)$, and one can show that $map(R[map_C(-,-)], F)\simeq \int_{c\in C}F$, the end of $F$. It follows that $Fun(C,Mod_R)^\omega$ is smooth if and only if $\int_{c\in C}: Fun(C^{op}\times C,Mod_R)\to Mod_R$ preserves filtered colimits.

Recall that ends are a special case limits over $Tw(C)$, so it suffices (but is not equivalent, as far as I know) for $\lim_{Tw(C)}$ to preserve filtered colimits. In this case, however, $Tw(\Delta^1)$ is a finite $\infty$-category, so this stronger condition is satisfied, and $Fun(\Delta^1,Perf(\mathbb F_p))$ is smooth, hence so is $R$.

Here's an example: consider the $\infty$-category $Fun(\Delta^1,Perf(\mathbb F_p))$. It has two canonical generators $A= \mathbb F_p\to \mathbb F_p$ and $B=\mathbb F_p\to 0$; and I claim that $R= End(A\oplus \Sigma B)$ is an example of what you're looking for (derived endomorphisms - more generally, everything here is "derived").

Note that $\hom(A,B) = \mathbb F_p$, $\hom(B,A)=0$, $\hom(A,A) = \mathbb F_p, \hom(B,B)= \mathbb F_p$.

In particular, $End(A\oplus\Sigma B) = \hom(A,\Sigma B)\oplus \hom(A,A)\oplus \hom(\Sigma B,A)\oplus \hom(\Sigma B,\Sigma B)$.

Note that $\hom(\Sigma B,\Sigma B)\simeq \hom(B,B)$, $\hom(\Sigma B,A) = 0$, so really $End(A\oplus \Sigma B)= \hom(A,A)\oplus \hom(B,B)\oplus \Sigma\hom(A,B)\simeq \mathbb F_p\oplus\mathbb F_p\oplus\Sigma\mathbb F_p$.

In particular, $End(A\oplus\Sigma B)$ is connective and perfect over $\mathbb F_p$, so it remains to argue that it is smooth. But smoothness is Morita invariant, i.e. it only depends on the category of modules.

Now, $A,B$ are generators of $Fun(\Delta^1,Perf(\mathbb F_p))$, so $Perf(End(A\oplus\Sigma B))\simeq Fun(\Delta^1,Perf(\mathbb F_p))$ by the Schwede-Shipley theorem (I'm using implicitly here that $Ind(C^{\Delta^1}) = Ind(C)^{\Delta^1}$ for any stable $C$), so it suffices to argue that $Fun(\Delta^1,Perf(\mathbb F_p))$ is smooth.

Now, for an indexing diagram $C$, $Fun(C,Mod_R)^\omega$ is smooth over $R$ if and only if $R[map_C(-,-)]$ is compact in $Fun(C^{op}\times C,Mod_R)$, and one can show that $map(R[map_C(-,-)], F)\simeq \int_{c\in C}F$, the end of $F$. It follows that $Fun(C,Mod_R)^\omega$ is smooth if and only if $\int_{c\in C}: Fun(C^{op}\times C,Mod_R)\to Mod_R$ preserves filtered colimits.

Recall that ends are a special case of limits over $Tw(C)$, so it suffices (but is not equivalent, as far as I know) for $\lim_{Tw(C)}$ to preserve filtered colimits. In this case, however, $Tw(\Delta^1)$ is a finite $\infty$-category, so this stronger condition is satisfied, and $Fun(\Delta^1,Perf(\mathbb F_p))$ is smooth, hence so is $R$.