Of course “several” in the title means $n$ strictly greater than 1 and the function is defined on some open subset of $\mathbb{C}^n$.
I tried to use the Weierstrass preparation theorem because it’s the only result on analysis of several variables I know but I couldn’t find a contradiction (I believe the answer is “no”).
The answer is indeed no, which might be surprising since the real analogue of this question has a different outcome ($f(x) = \|x\|^2$ is smooth and only has one zero).
The crucial result in this proof is Hartog's Theorem:
Let $n \geq 2$ and $\Omega \subset \mathbb C^n$ be open. Suppose $K \subset \Omega$ is compact and $\Omega \setminus K$ is connected. Then any holomorphic function on $\Omega \setminus K$ can be extended to all of $\Omega$.
Remark: This is not true in one variable: The function $z \mapsto e^{1/z}$ cannot be extended to all of $\mathbb C$ because it has an essential singularity at zero.
We are now going to prove that no holomorphic function of several variables has just one zero.
Toward a contradiction, let $\Omega \subset \mathbb C^n$ be a (connected) domain and let $f: \mathbb{\Omega} \rightarrow \mathbb C$ be holomorphic with $f(z) = 0$ if and only if $z = z_0$. Then the function $1/f$ is holomorphic in $\Omega \setminus \{z_0\}$. By Hartog's Theorem, $1/f$ admits a holomorphic extension to $\Omega$. But then, $f(z_0) \cdot 1/f(z_0) = 1$ by continuity of the product of continuous functions, which is a contradiction to $f(z_0) = 0$.
The answer is no for a function defined on an open subset of the plane. It actually comes down to the following algebra fact :
Let $A$ be a normal ring, then $$A=\cap_{ht(\mathfrak{p})=1} A_{\mathfrak{p}}.$$
where if $\mathfrak{p}$ is a prime ideal of $A$, the height of $\mathfrak{p}$ denoted by $ht(\mathfrak{p})$ is the maximal number of prime ideals contained in $\mathfrak{p}$ that form a strictly increasing chain $$\mathfrak{p_1} \subsetneq\cdots\subsetneq \mathfrak{p_s}\subsetneq \mathfrak{p}.$$
And $A_{\mathfrak{p}}$ is the localization of $A$ with respect to $\mathfrak{p}$.
A proof can be found in Matsumura's "Commutative Algebra".
Now, you may suppose that your open subset of the plane is connected, hence the ring of holomorphic functions on $U$ is normal (this is an exercise). Denote it by $A$ and denote by $K$ its fraction field i.e., the field of meromorphic functions on $U$.
My claim is that if $f$ is holomorphic and has only one point as zero locus, then $1/f$ is holomorphic, which yealds to a contradiction.
But then I only have to prove that $1/f$ is in all the $A_{\mathfrak{p}}$ for all primes of height one.
To do this, I say that the elements of a certain prime of height one of $A$ are exactly the holomorphic functions vanishing on a certain codimension $1$ subset $Z$ of $U$ (this is the analytic version of the Nullstellensatz, again a good exercise).
Therefore $1/f$ is not in any $A_{\mathfrak{p}}$.
The vanishing set of an holomorphic function $\Bbb{C}^{n}\to\Bbb{C}$ is an union of analytic hypersurfaces (locally biholomorphic to $\Bbb{C}^{n-1}$)
Use the n-polydisk Cauchy integral formula to show holomorphic implies analytic. Let $$f(w,z)=F_w(z)=\sum_n g_n(w) z^n, \qquad w\in \Bbb{C}^{n-1},z\in \Bbb{C}$$ be analytic non-constant and $f(0,0)=0$. The $g_n$ are analytic with a common non-zero radius of convergence.
If $F_0'(0) \ne 0$ it is easy : $F_w'(0)\ne 0$ for $w$ small thus with $r$ small enough from the residue thoerem $$h(w) = \frac1{2i\pi} \int_{|z|=r} \frac{z F_w'(z)}{F_w(z)}dz, \qquad f(w,h(w))=0$$
Otherwise $F_0'(0) =0$, assume $F_0^{(k)}(0)\ne 0$ then from the argument principle the number of zeros of $F_w$ near $z=0$ is the number of times the curve $F_w(r e^{it}),t\in [0,2\pi]$ encloses $0$, for $w$ small it is $\le k$ and it is continuous in $w$.
There may be several hypersurfaces passing through $w=0$ but for $w\ne 0$ small the hypersurfaces separate and replacing $\int_{|z|=r}$ by $\int_\gamma$ we get the equations of our hypersurfaces.
The remaining case is $F_0$ identically zero : thus $f(z,w) = z^m f_2(z,w)$ and if $f_2(0,0)=0$ then $f_2$ satisfies the previous hypothesis.
Suppose for simplicity that $f$ is holomorphic in an open set $U\subset \mathbb C^2$ with $(0,0)\in U.$ Assume that $f$ has an isolated zero at $(0,0).$ Then we can choose $r>0$ such that $f$ is holomorphic in $D(0,2r)^2$ and $f\ne 0$ in $D(0,2r)^2\setminus \{(0,0)\}.$
For $w\in D(0,r)$ define
$$g(w)=\frac{1}{2\pi i}\int_{|z|=r} \frac{D_zf(z,w)}{f(z,w)}\,dz.$$
The argument principle shows that $g(0)$ is a nonzero integer $m,$ while $g(w)= 0$ for $0<|w|<r.$ Note also that $g$ is a continuous function of $w.$ Thus $g(0) = m = \lim_{z\to 0} g(w)=0,$ which is a contradiction.
