This question already has an answer here:
As Jyrki Lahtonen mentioned in the comment above, an induction argument works here. Let me sketch the argument in the base case when $n=2$ and then you can work from here.
Suppose you have a polynomial $F(X_1, X_2) \in K[X_1, X_2]$ with $\deg{F} \geq 1$. We can write $F$ as a polynomial in $X_1$ with coefficients in $K[X_2]$ as follows
$$
F(X_1, X_2) = \sum_{i = 0}^d c_i(X_2) X_1^{i}
$$
where $c_i(X_2) \in K[X_2]$, and $d$ is the degree of $F$ in the variable $X_1$. Now, it may happen that $d = 0$ so that your polynomial only involves the variable $X_2$, but then in that case since $F$ has at least one zero $b \in K$ since the field is algebraically closed, then any point $(a, b) \in K^2$ would automatically also be a zero of $F$, so that you’ll have infinitely many zeros.
On the other case, if $d \geq 1$, then you can look at the polynomial $c_d(X_2) \in K[X_2]$ which is the coefficient of the highest degree term in $X_1$. This polynomial $c_d(X_2) \neq 0$, and therefore has finitely many zeros in $K$ (if any, because it may be constant). Therefore if the set of zeros of $c_d(X_2)$ is $\{ b_1, \dots , b_r \}$, then for any value of $b \in K \setminus \{ b_1, \dots , b_r \}$ we have $c_d(b) \neq 0$.
Therefore the polynomial $F_b(X_1) := F(X_1, b) = \sum_{i = 0}^d c_i(b) X_1^{i} \in K[X_1]$ is a degree $d$ polynomial in one variable, so it has $d$ zeros $a_1, \dots a_d$. Therefore the points $(a_i, b)$ are zeros of the polynomial $F(X_1, X_2)$ and therefore since there are infinitely many distinct possibilities for $b$, then we can conclude again that $F(X_1, X_2)$ has infinitely many zeros.
From here you can do the induction step.