No. Consider $K$ a field and $R=K[x_n:n\ge 0]/(x_n^2:n\ge 0)$.

Clearly $R$ is local and its nilradical equals its unique maximal ideal, so it has Krull dimension zero. It has the properly descending sequence of ideals $R\supset (x_0)\supset (x_0x_1)\supset\cdots$.

But it has ACC on principal ideals.

Indeed, consider $(a_k)_{k\ge 0}$ such that $a_{k+1}$ divides $a_k$ for all $k$ (say $a_k=a_{k+1}d_k$); we have to show that $a_k$ divides $a_{k+1}$ for large enough $k$. This is clear if some $a_k$ is invertible, hence assume they're all in the maximal ideal. Also the case all $a_k$ zero being trivial, we can suppose (extracting if necessary) $a_0\neq 0$.

Let $I$ be the (nonempty finite) set of $n$ such that $x_n$ occurs in $a_0$. We consider the linear decomposition $R=R_I\oplus J_I$, where $R_I$ is the subring $K[x_i:i\in I]$ and $J_I$ is the ideal generated by the $x_n$ for $n\notin I$. Then the projection $R\to R_I$ (with respect to this decomposition) is a $K$-algebra homomorphism.

Write $a_k=b_k+c_k$ and $d_k=e_k+f_k$ in the above decomposition. Then $b_k=b_{k+1}e_k$ for all $k$. Since it belongs to the noetherian subring $R_I$, for $k$ large enough, $b_k$ is a nonzero scalar multiple of $b_{k+1}$. Up to rescaling, we suppose henceforth that all $b_k$ are equal for large enough $k$, say to $b$. Since $b$ divides $b_0=a_0\neq 0$, we have $b\neq 0$.

So $be_k=b$ for all large enough $k$, that is, $b(e_k-1)=0$. Since $b\neq 0$, this implies that $e_k-1$ is a zero divisor, and hence belongs to the maximal ideal, which implies that $e_k$ is invertible. So $d_k$ is invertible, and hence the sequence $(a_k)$ is stationary.