Fix a field $K$. The following is a basic fact about central simple algebras.
Theorem. Let $A$ be a central simple $K$-algebra, and let $B$ be an arbitrary $K$-algebra. Any two-sided ideal $\mathfrak{a} \subset A \otimes_K B$ has the form $A \otimes_K \mathfrak{b}$ where $\mathfrak{b} \subset B$ is a two-sided ideal. (In particular, $\mathfrak{b} = \mathfrak{a} \cap B$.)
I'm looking for the analog of this statement replacing $A$ with a field extension of $K$.
Theorem. Let $L/K$ be a field extension, and let $B$ be an arbitrary $K$-algebra. Any two sided-ideal $\mathfrak{a} \subset L \otimes_K B$ has the form...?
I do not believe that the precise analog holds, that is, not every ideal has the form $L \otimes_K \mathfrak{b}$ for some ideal $\mathfrak{b} \subset B$. As a counterexample, let $L= \mathbb{C}$ and $A = \mathbb{R}[x]$, and then consider the tensor product $\mathbb{C} \otimes_\mathbb{R} \mathbb{R}[x] \cong \mathbb{C}[x]$. The principal ideal $(x+i) \subset \mathbb{C}[x]$ does not come from an ideal of $\mathbb{R}[x]$. This example hints that ideals of $L \otimes_K B$ should have the form $$ \bigoplus_{i=1}^{n} L \otimes_K \mathfrak{b}_i $$ where $\mathfrak{b}_i \subset B$ is an ideal, and $n \le [L:K]$.
Possibly there is not much to say in this generality, I'm also happy if something can be said assuming $L/K$ is finite and separable, or even Galois. If assumptions like $\operatorname{char}(K) = 0$ are involved, that's fine too. This seems like a relatively simple thing, but I don't see it written down anywhere in this form.
Your conjecture, if I understand it correctly, is false and your counterexample is a counterexample to it. Here is another even worse counterexample: let $K = \mathbb{F}_p(t)$ and let $L = B$ be the purely inseparable extension $K[x]/(x^p - t)$. Then $B$ is a field so has no nontrivial ideals but
$$L \otimes_K B \cong L[x]/(x^p - t) \cong L[x]/(x - \sqrt[p]{t})^p$$
even has nontrivial nilradical!
In the case that $L/K$ is a finite Galois extension this is related to Galois descent; the case of $B = K[x]$ is a nice concrete example. The ideals of $K[x]$ are principal ideals $(f(x))$ generated by some polynomial and they factor into a product of prime ideals corresponding to the irreducible factors of $f(x)$. After tensoring with $L$ some irreducible polynomials become reducible so the corresponding prime ideals factor further; any new factors arising in this way are new ideals in $L[x]$ which are not extensions of scalars of ideals in $K[x]$.
On the other hand, what we can say is that the Galois group $G$ acts on ideals, and an ideal in $L[x]$ is the extension of scalars of an ideal in $K[x]$ iff it's closed under the action of the Galois group (this will generally be true for an arbitrary $B$, by Galois descent for subspaces of vector spaces). What this says algebraically is that a monic polynomial in $L[x]$ has coefficients in $K[x]$ iff it's fixed by the Galois action (this is precisely the definition of a Galois extension), and what this says geometrically is that closed subschemes of the affine line $\mathbb{A}^1/K$ correspond to closed subschemes of the affine line $\mathbb{A}^1/L$ which are Galois invariant; in particular, closed points of $\mathbb{A}^1/K$ correspond to orbits of the Galois action on closed points of $\mathbb{A}^1/L$.
Tensoring from $K$ up to $L$ has the effect of "breaking apart" these orbits so that their individual constituents are available; in your example where $K = \mathbb{R}, L = \mathbb{C}$ the maximal ideal $(x^2 + 1) \subset \mathbb{R}[x]$ breaks apart into two maximal ideals $(x + i), (x - i) \subset \mathbb{C}[x]$ reflecting the breaking apart of the Galois orbit $\{ i, -i \}$ into its two points.