= Solution
Put $\alpha=\sqrt{m_1}$, $\beta=\sqrt{m_2}$, and
$$
\omega=\frac{1+\beta}{2}.
$$
The proposed <integral basis for coprime biquadratic radicands congruent to two and one> is
$$
\boxed{\{1,\alpha,\omega,\alpha\omega\}}.
$$
Each entry is an <algebraic integer>, by the <ring of integers of a quadratic field> formula and closure under products. Their coefficient <matrix> in $1,\alpha,\beta,\alpha\beta$ has <determinant> $1/4$, so they are rationally <linearly independent>. It remains to show that they span every <algebraic integer>.
Write $\gamma=(a+b\alpha+c\beta+d\alpha\beta)/2$ initially with rational $a,b,c,d$. The <field traces> to $\mathbb Q(\alpha)$ and $\mathbb Q(\alpha\beta)$ are $a+b\alpha$ and $a+d\alpha\beta$. Since $m_1\equiv m_1m_2\equiv2\pmod4$, the corresponding <integral basis> in each subfield gives $a,b,d\in\mathbb Z$. The <field trace> to $\mathbb Q(\beta)$ is $a+c\beta$. Membership in $\mathbb Z[\omega]$ says that $2a,2c$ are <integers> of equal parity. As $a$ is already an <integer>, this forces $c\in\mathbb Z$ too.
Now the <field norm> to $\mathbb Q(\beta)$ is
$$
\frac A4+\frac{ac-m_1bd}{2}\beta,
\qquad A=a^2+m_2c^2-m_1b^2-m_1m_2d^2.
$$
By the <ring of integers of a quadratic field> formula, $A/2$ is an <integer> and has the same parity as $ac-m_1bd$. Thus $A$ is even, giving $a\equiv c\pmod2$, and
$$
A\equiv2ac-2m_1bd\equiv2ac\pmod4.
$$
Using $m_1\equiv2$, $m_2\equiv1$ and $a\equiv c\pmod2$, this becomes
$$
0\equiv A-2ac\equiv(a-c)^2-2(b^2+d^2)
\equiv-2(b^2+d^2)\pmod4.
$$
Hence $b\equiv d\pmod2$. Finally,
$$
\gamma=\frac{a-c}{2}+\frac{b-d}{2}\alpha+c\omega+d\alpha\omega,
$$
an expression with <integer> coefficients. This proves the asserted <integral basis> without assuming a general theorem about products of elements of an <integral basis>.
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