It can't even be embedded bilipschitz (or even quasi-isometrically) into a Hilbert space. A proof using basic unitary representation theory was given in a paper of mine with Tessera and Valette (GAFA 2007).

The proof works as follows. Let $u$ be such an quasi-isometric embedding (of a compactly generated locally group $G$ for the moment). Then $f:(x,y)\mapsto \|u(x)-u(y)\|$ is a conditionally negative definite kernel on $G$, with the property that
$$ c^{-1}|x^{-1}y|-C\le \|f(x,y)\|\le c|x^{-1}y|+C$$
for all $x,y$ and some constants $c>0$ and $C$, and $|\cdot|$ denotes the distance (left-invariant) to 1.

Assuming that $G$ is amenable, in a first step change $f$ to a finite perturbation that is uniformly continuous, average $f$ along left cosets, defining $g(x,y)=\int_{k\in G}f(kx,ky)dm$, where $m$ is an invariant mean; then $g$ is continuous (by some little argument, unnecessary if $G$ is discrete, notably the discrete Heisenberg group), and $g$ is a left-invariant conditionally negative definite kernel: $g(x,y)=h(x^{-1}y)$ for some continuous conditionally negative definite function $h$. So, there is an affine isometric representation of $G$ on a Hilbert space such that $h(x)=\|x\cdot 0\|^2$ for all $x\in G$.

The remainder is some stuff on isometric affine action in Hilbert spaces. An old (and quite easy) result of Guichardet is that if $G$ is nilpotent and $\pi$ a unitary/orthogonal representation with no nonzero invariant vectors, then $\overline{H^1}(G,\pi)=0$. Then a simple argument shows that for $G$ arbitrary and an arbitrary unitary/orthogonal representation, every cocycle that is zero in reduced cohomology, has sublinear growth.

So now for $G$ nilpotent as above, we decompose the representation as invariant $\oplus$ (orthogonal of invariant vectors). This decomposes the cocycle in $b+b'$, where $b'$ grows sublinearly. But $b$ being a cocycle of a trivial representation, it's a homomorphism and in particular is zero on $[G,G]$. Hence if $\overline{[G,G]}$ is non-compact, $b+b'$ cannot grow linearly.