This reminds me of what someone once said to me:

The geometers can always take a hyperplane section. We can't!

The purpose of this post is to analyze the question to see how close it is to Bertini's theorem (it is not obvious to me). The statement is immediately equivalent to: one can always find a nonzero prime ideal $P \subseteq m_1\cap m_2$ (since if we can, then induction on dimension proves the original).

Now, how can such $P$ exist? Let $U_1 = R-m_1$ and $U_2=R-m_2$. Let $U=\{xy \| x\in U_1, y\in U_2\}$. $U$ is multiplicative and our prime $P$ obviously just has to avoid $U$. So $P$ exists unless the localization $U^{-1}R$ has dimension $0$. But it is a domain, so we have to make sure $U^{-1}R$ is not a field. That statement is equivalent to the existence of some element $f\in R$ such that $f$ does not become an unit in $U^{-1}R$. In other words:

there are no $a,b \in U$ such that $af=b$.

Since $b$ is itself a product of elements in $U_1,U_2$, our condition is obvious if $fR$ is a prime ideal and $f\in m_1\cap m_2$. But it is technically weaker, although not clear to me by how much. Note that we do not require $f$ to be linear, which is common for Bertini's type statement.

This analysis would seem to rule out certain clever arguments. But may be one can find some!

For "algebraic" version of Bertini's theorem, I will look at the reference given here. Also, how to rescue Bertini over finite fields using hypersurface instead of hyperplane (which suits your purpose), looks here.

This reminds me of what someone once said to me:

The geometers can always take a hyperplane section. We can't!

The purpose of this post is to analyze the question to see how close it is to Bertini's theorem (it is not obvious to me). The statement is immediately equivalent to: one can always find a nonzero prime ideal $P \subseteq m_1\cap m_2$ (since if we can, then induction on dimension proves the original).

Now, how can such $P$ exist? Let $U_1 = R-m_1$ and $U_2=R-m_2$. Let $U=\{xy \| x\in U_1, y\in U_2\}$. $U$ is multiplicative and our prime $P$ obviously just has to avoid $U$. So $P$ exists unless the localization $U^{-1}R$ has dimension $0$. But it is a domain, so we have to make sure $U^{-1}R$ is not a field. That statement is equivalent to the existence of some element $f\in R$ such that $f$ does not become an unit in $U^{-1}R$. In other words:

there are no $a,b \in U$ such that $af=b$.

Since $b$ is itself a product of elements in $U_1,U_2$, our condition is obvious if $fR$ is a prime ideal and $f\in m_1\cap m_2$. But it is technically weaker, although not clear to me by how much. Note that we do not require $f$ to be linear, which is common for Bertini's type statement.

This analysis would seem to rule out certain clever arguments. But may be one can find some!

For "algebraic" version of Bertini's theorem, I will look at the reference given here. Also, how to rescue Bertini over finite fields using hypersurface instead of hyperplane (which suits your purpose), looks here.