suggestion (requires: composition of smooth maps is smooth):
Suppose that for all $f:N\mathbb \rightarrow R$, $f\circ F$ is smooth.
Let $y:V\rightarrow \mathbb R^n$ be a chart for $N$. Then $y^i\circ F$ is smooth for all $i$ $\Rightarrow$ $y\circ F$ is smooth
Let $x:U\rightarrow \mathbb R^m$ be a chart for $M$, $U\subset F^{-1}(V)$. Then the map $(y\circ F) \circ x^{-1}$ is smooth.
EDIT: there is a glitch (see comment above) - $y^i$ is not defined on $N$, so the argument does not work! I think there is no way to avoid bump functions.
First one has to show: For all $V\subset N$ open and smooth $f:V \rightarrow \mathbb R$ the map $ f \circ (F|_{F^{-1}(V)}) $ is smooth. If $p\in V$ is a point, choose a smooth bump function $u\,$ s.t. $\text{supp}(u) \subset V$ and $u=1$ in an open neighbourhood of $p$, then $uf$ extends to a smooth function on $N$ and so on.