Let $A_\ell$ be the number of monomials of degree $n$ on $\ell$ variables, which involve all $\ell$ variables and satisfy the condition (this will necessarily be 0 for $\ell>n$). The number of monomials involving all $\ell$ variables is $\binom{n-1}{\ell-1}$ by stars-and-bars. The number of monomials involving all $\ell$ variables at least twice (the invalid monomials), dividing by $x_1\cdots x_\ell$, is $\binom{n-\ell-1}{\ell-1}$. Thus $A_\ell=\binom{n-1}{\ell-1}-\binom{n-\ell-1}{\ell-1}$.
Each monomial is supported on a unique subset of the variables. For a fixed subset of size $\ell$, the monomials supported there are counted by $A_\ell$. There are $\binom{k}{\ell}$ subsets of size $\ell$. So if $N_k$ is the answer to the problem, I believe we have the formula
$$N_k=\sum_{0\leq \ell\leq k} A_\ell \binom{k}{\ell}=\sum_{0\leq \ell\leq k}\binom{n-1}{\ell-1}\binom{k}{\ell}-\binom{n-\ell-1}{\ell-1}\binom{k}{\ell}$$ $$=\binom{n+k-1}{n}-\sum_{0\leq \ell\leq k}\binom{n-\ell-1}{\ell-1}\binom{k}{\ell}$$
[Edit: I now see that this argument was already given by Vladimir Dotsenko. There seems to be some disagreement about his answer though, so I will leave this here as independent confirmation.]
