I think more is true. Let $M^n$ be a closed connected simply connected manifold of dimension $\ge 3$ not homeomorphic to $S^n$. Let $p,q\in M$ be two distinct points. Then $M\setminus\{p,q\}$ is not the universal cover of a closed manifold.
Suppose it is and $M=\tilde N$ is the universal cover of a closed manifold $N^n$. As HJRW suggested in a comment $\pi_1(N)$ must have two ends and hence must be virtually $\mathbb Z$. By passing to a finite index subgroup we can assume it's $\mathbb Z$. Since the action of $\pi_1(N)$ on $\tilde N$ must permute the ends we can again assume by passing to finite index subgroup that it fixes the ends. Then it can be extended to a continuous action on $M$ fixing $p$ and $q$. Take a peripheral $X_1=S^{n-1}$ around $p$. Let $g$ be a generator of $\pi_1(N)\cong \mathbb Z$. Let $\bar D^n$ be the disk centered at $p$ with boundary $X_1$. By possibly changing $g$ to $g^{-1}$ we can assume that $X_2=g(X_1)$ is a sphere in $\bar D^n$. By the annulus theorem the region $W_1$ between $X_1$ and $X_2$ is homeo to $S^{n-1}\times [0,1]$. Then $W_2=g(W_1)$ is a manifold with boundary glued to $W_1$ along $X_2$. Then $g^2(W_2)$ will be another copy of $S^{n-1}\times [0,1]$ glued to $W_2$ along the second piece of the boundary. Continuing we get a copy of $\mathbb S^{n-1}\times \mathbb R$ sitting in $\tilde N$ with the standard action of $\mathbb Z$. Passing to the quotient by $\mathbb Z$ this gives an embedding of $\mathbb S^{n-1}\times \mathbb S^1$ into $N$ which must be a homeomorphism since these are connected closed manifolds of the same dimension. Then $\tilde N$ is homeo to $ \mathbb S^{n-1}\times \mathbb S^1$ and $M$ is homeo to $\mathbb S^n$.