Consider I have a vector space $V$ with inner product and a bilinear map $b : V \times V \to V$
i) such that if $z = b(u,v)$ for two $u,v \in V$, then $$ z \perp u \quad \mbox{ and } \quad z \perp v. $$ ii) if $u, v \in V$ are perpendicular, i.e. $u \perp v$, then $$ ||b(u,v)|| = ||u||||v||. $$ These definitions are motivated by an axiomatic introduction of the cross product, see here.
Now I want to show that for every $v \in V$ we have $b(v,v) = 0$.
If $V$ is finite-dimensional, then this follows by the fact that in an $n$-dimensional space, if $\{v_1, \ldots, v_m\}$ are orthogonal and non-zero, then $m \le n$ (because orthogonality of non-zero vectors implies linear independence). For if $v \in V, v \ne 0$ then the vectors $$ M = \{ v, b(v,v), b(v,b(v,v)), \ldots, b(v, b(v, \ldots, b(v,v))) \} $$ are all orthgonal, to simplify notation suppose we have a $n = 4$ dimensional space. Then $$ b(v, b(v, b(v, b(v, v))) = 0 $$ which implies by ii) \begin{align*} 0 & = ||b(v, b(v, b(v, b(v,v)))|| \\ & = ||v|| ||b(v, b(v, b(v,v)))|| \\ & = ||v|| ||v|| ||b(v, b(v,v))|| \\ & = ||v|| ||v|| ||v|| ||b(v,v)|| \end{align*} which implies $||b(v,v)|| = 0$, because $v \ne 0$, which implies that $b(v,v)$ is the zero vector.
But does this also hold if $V$ is infinite-dimensional, if not can you give an example were it fails?
If such a bilinear form exists, it must satisfy $b(v,v) = 0$.
Let $w = b(v,v)$, and consider $b(v, v+w)$. By property i) and bilinearity, \begin{align*} 0 &= \langle v+w, b(v,v+w)\rangle\\ &= \langle v, b(v,v+w)\rangle + \langle w, b(v,v+w)\rangle\\ & = \langle w, b(v,v+w)\rangle\\ &= \langle w, b(v,v)\rangle + \langle w, b(v,w)\rangle\\ &= \langle w, b(v,v)\rangle\\ &= \|w\|^2 \end{align*} and $w=0$.
Note that the above requires that $V$ be an inner product space; if you want to generalize the cross product to "all" of infinite-dimensional Euclidean space you will first need to write down a definition of $\perp$ for vectors that are not in $\ell^2$.