This is an attempt to complete Tyler's argument. We first note that $K^0(S^5)=\mathbb Z$ (note there is a parity difference, $K^*(S^n)$ is always $\mathbb Z^2$ but for $n$ odd we have $K^1(S^n)=\mathbb Z$ and hence $K^0(S^n)=\mathbb Z$). This means that every topological vector bundle on $S^n$ is topologically stably trivial. Let now $E$ be an algebraic vector bundle on $S^n$, i.e., an f.g. projective module over $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$, of rank $k$. As it is topologically stably trivial that means that there are continuous sections $f_1,\dots,f_k$ of some $E\bigoplus R^m$ trivialising it, i.e., form a basis at each fibre. Now, being a trivialisation is an open condition under the sup norm (with respect to some metric on the vector bundle to be precise) so if we can show that the algebraic sections of any vector bundle $F$ are dense in the space of continuous sections we get that it is also algebraically trivial. However, picking a $G$ such that $F\bigoplus G$ is trivial reduces this to showing that $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$ is dense in the ring of continuous (real-valued) functions on $S^n$ but this follows from the Stone-Weierstrass theorem.

Addendum: I made a small thinko, I implicitly use that an element of $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$ that is invertible as continuous function on the sphere is invertible in the ring which of course is not true (it can be zero on a complex point of the sphere). Hence, it is necessary to consider the localisation where all such functions are inverted. This still gives a Noetherian ring however.

This is an attempt to complete Tyler's argument. We first note that $KO^0(S^5)=\mathbb Z$ (note this true for all spheres of dimension $\equiv 5,6,7 \bmod 8$). This means that every topological vector bundle on $S^5$ is topologically stably trivial. Let now $E$ be an algebraic vector bundle on $S^n$, i.e., an f.g. projective module over $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$, of rank $k$. As it is topologically stably trivial that means that there are continuous sections $f_1,\dots,f_k$ of some $E\bigoplus R^m$ trivialising it, i.e., form a basis at each fibre. Now, being a trivialisation is an open condition under the sup norm (with respect to some metric on the vector bundle to be precise) so if we can show that the algebraic sections of any vector bundle $F$ are dense in the space of continuous sections we get that it is also algebraically trivial. However, picking a $G$ such that $F\bigoplus G$ is trivial reduces this to showing that $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$ is dense in the ring of continuous (real-valued) functions on $S^n$ but this follows from the Stone-Weierstrass theorem.

Addendum: I made a small thinko, I implicitly use that an element of $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$ that is invertible as continuous function on the sphere is invertible in the ring which of course is not true (it can be zero on a complex point of the sphere). Hence, it is necessary to consider the localisation where all such functions are inverted. This still gives a Noetherian ring however.

Addendum 1: I confused real and complex K-theory in my initial description which is now fixed. I also (unsurprisingly) found a reference for the arguments in Bochnak, Coste, Roy: Real algebraic geometry, Cap 12.

This is an attempt to complete Tyler's argument. We first note that $K^0(S^5)=\mathbb Z$ (note there is a parity difference, $K^*(S^n)$ is always $\mathbb Z^2$ but for $n$ odd we have $K^1(S^n)=\mathbb Z$ and hence $K^0(S^n)=\mathbb Z$). This means that every topological vector bundle on $S^n$ is topologically stably trivial. Let now $E$ be an algebraic vector bundle on $S^n$, i.e., an f.g. projective module over $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$, of rank $k$. As it is topologically stably trivial that means that there are continuous sections $f_1,\dots,f_k$ of some $E\bigoplus R^m$ trivialising it, i.e., form a basis at each fibre. Now, being a trivialisation is an open condition under the sup norm (with respect to some metric on the vector bundle to be precise) so if we can show that the algebraic sections of any vector bundle $F$ are dense in the space of continuous sections we get that it is also algebraically trivial. However, picking a $G$ such that $F\bigoplus G$ is trivial reduces this to showing that $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$ is dense in the ring of continuous (real-valued) functions on $S^n$ but this follows from the Stone-Weierstrass theorem.

This is an attempt to complete Tyler's argument. We first note that $K^0(S^5)=\mathbb Z$ (note there is a parity difference, $K^*(S^n)$ is always $\mathbb Z^2$ but for $n$ odd we have $K^1(S^n)=\mathbb Z$ and hence $K^0(S^n)=\mathbb Z$). This means that every topological vector bundle on $S^n$ is topologically stably trivial. Let now $E$ be an algebraic vector bundle on $S^n$, i.e., an f.g. projective module over $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$, of rank $k$. As it is topologically stably trivial that means that there are continuous sections $f_1,\dots,f_k$ of some $E\bigoplus R^m$ trivialising it, i.e., form a basis at each fibre. Now, being a trivialisation is an open condition under the sup norm (with respect to some metric on the vector bundle to be precise) so if we can show that the algebraic sections of any vector bundle $F$ are dense in the space of continuous sections we get that it is also algebraically trivial. However, picking a $G$ such that $F\bigoplus G$ is trivial reduces this to showing that $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$ is dense in the ring of continuous (real-valued) functions on $S^n$ but this follows from the Stone-Weierstrass theorem.

Addendum: I made a small thinko, I implicitly use that an element of $\mathbb R[x_0,\dots,x_n]/(x_0^2+\cdots+x_n^2-1)$ that is invertible as continuous function on the sphere is invertible in the ring which of course is not true (it can be zero on a complex point of the sphere). Hence, it is necessary to consider the localisation where all such functions are inverted. This still gives a Noetherian ring however.