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The answer is no. Take any projective manifold $X$ mapping $\pi\colon X\rightarrow\mathbb P^n$ to such that $\pi_*\mathcal O_X=\mathcal O_{\mathbb P^n}$ and let $L$ be $\pi^*\mathcal O(1)$. Then $H^0(X,L^{\otimes m})=H^0(\mathbb P^n,\mathcal O(m))$ Examples are $X=Y\times\mathbb P^n$ or a blowing up of a closed smooth subvariety of $\mathbb P^n$.

I do not know if some power of $L$ is necessarily without base points (in which case $L$ itself is). In any case we can blow up $X$ such that there is a map to $\mathbb P^n$. The line bundle that gives that mapping may I guess have more sections.

Addendum: I was really rambling in the last paragraph. What I meant was that suppose we only know that $\bigoplus_nH^0(X,L^{\otimes n})$ is a polynomial ring. Does that mean that $L$ is base-point free? It may very well be true but I don't see it at the moment.

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The answer is no. Take any projective manifold $X$ mapping $\pi\colon X\rightarrow\mathbb P^n$ to such that $\pi_*\mathcal O_X=\mathcal O_{\mathbb P^n}$ and let $L$ be $\pi^*\mathcal O(1)$. Then $H^0(X,L^{\otimes m})=H^0(\mathbb P^n,\mathcal O(m))$ Examples are $X=Y\times\mathbb P^n$ or a blowing up of a closed smooth subvariety of $\mathbb P^n$.

I do not know if some power of $L$ is necessarily without base points (in which case $L$ itself is). In any case we can blow up $X$ such that there is a map to $\mathbb P^n$. The line bundle that gives that mapping may I guess have more sections.