Let $k$ be an algebraically closed field and the field $K$ contains $k$. I am trying to prove that if $F_1,...,F_m\in k[x_1,...,x_n]$ and the system of polynomial equations $F_1=0,...,F_m=0$ has a solution in $K$ then it has a solution in $k$.
Denote by $I$ the ideal $(F_1,...,F_m)$.
By Hilbert's Nullstellensatz, weak form, I have that if $V(I)=\emptyset$ then $1\in I$. But $1$ obviously does not have zeros in $K$.
Is this argument correct?
Yes, this argument is correct. To put it in a slightly different form, you have an embedding $k \hookrightarrow K$, so $\varphi: k[x_1,..., x_n] \hookrightarrow K[x_1,...,x_n]$. If you have no solution in $k$ then $1 \in I$, so $1 = \varphi(1) \in \varphi_*I = (\varphi(F_1),...,\varphi(F_n))$ and so you have no solution in the bigger field.