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This is based on Anonymous answer.

Theorem. Let $X_0$ be a strongly zero-dimensional Tychonoff space, $Y\subseteq \beta X_0\setminus X_0$ a locally compact scattered space. Then $X = \beta X_0\setminus Y$ is such that $\beta X\setminus X = Y$ and $X$ has only zero-dimensional compactifications.

Remark. In Anonymous example, we have $Y = N$ and $X_0 = \mathbb{N}$, and all the assumptions hold. Clearly $Y$ is compact iff $X$ is locally compact.

Proof:

Suppose $K$ is a compactification of $X$ and let $S\subseteq K$ be a compact connected set with more than two points. Note that $\beta X = \beta X_0$ since $X_0\subseteq X \subseteq \beta X_0$ and so there exists a surjective map $\beta X_0\to K$ which maps $X$ to itself and $Y$ to $K\setminus X$.

Since $\beta X_0\setminus \text{cl}_{\beta X_0} Y \subseteq K$ is open, given $p\in S \cap (\beta X_0\setminus \text{cl}_{\beta X_0} Y)$ and $q\in S, q\neq p$, there exists a clopen compact subset $U$ of $\beta X_0\setminus \text{cl}_{\beta X_0} Y$ with $p\in U, q\notin U$, which is impossible. So that we need to have $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y\cup K\setminus X$. Let $L = \text{cl}_{\beta X_0} Y \setminus Y\cup K\setminus X$, $L$ compact.

Since $Y$ is locally compact, $\text{cl}_{\beta X_0} Y \setminus Y$ is closed in $L$, so $K\setminus X$ is open in $L$. The map $Y\to K\setminus X$ is perfect and so by this answer by K. P. Hart, $K\setminus X$ is open and zero-dimensional. If now $p\in S\cap (K\setminus X)$ and $q\in S, p\neq q$, then we can take a clopen compact subset $U$ of $L$ with $p\in U, q\notin U$, so as previously we can assume $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$.

Since $\text{cl}_{\beta X_0} Y \setminus Y$ is subspace of zero-dimensional space, it's itself zero-dimensional. And so there can't be a connected $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$ with more than two points.

It follows that $K$ is totally disconnected and compact, and so zero-dimensional. $\square$

Example. If $Y$ is a locally compact, scattered space, then there exists a space $X$ with only zero-dimensional compactifications and $\beta X\setminus X\cong Y$.

Proof: Let $\alpha > 0$ be a non-limit ordinal with $|\beta Y| < \aleph_\alpha$ and $X_0 = \beta Y\times \omega_\alpha$. Then $\beta X_0 = \beta Y \times (\omega_\alpha + 1)$. The space $X = \beta X_0\setminus (Y\times \{\omega_\alpha\})$ is such that $\beta X\setminus X \cong Y$. $\square$

Corollary. Let $X$ be a strongly zero-dimensional Tychonoff space with $\beta X\setminus X$ locally compact. Then the following are equivalent:

1. $X$ has only zero-dimensional compactifications.

2. $\beta X\setminus X$ is scattered.

Proof: 2 implies 1 from the theorem.

Suppose $X$ has only zero-dimensional compactifications. Then $\beta X\setminus X$ is zero-dimensional as a subspace of $\beta X$. If $\beta X\setminus X$ weren't scattered, then we would find a compact perfect set $A\subseteq \beta X\setminus X$ as in K. P. Hart answer and so there would exist a surjection $A\to [0, 1]$. Since $\beta X\setminus A$ is locally compact, from Engelking theorem 3.5.13 there exists a compactification $Z$ of $\beta X\setminus A$ with remainder homeomorphic to $[0, 1]$. Its also a compactification of $X$, and $Z$ is not zero-dimensional. The contradiction shows that such set $A$ cannot exist, and so $\beta X\setminus X$ is scattered. $\square$

This is based on Anonymous answer.

Theorem. Let $X_0$ be a strongly zero-dimensional Tychonoff space, $Y\subseteq \beta X_0\setminus X_0$ a locally compact scattered space. Then $X = \beta X_0\setminus Y$ is such that $\beta X\setminus X = Y$ and $X$ has only zero-dimensional compactifications.

Remark. In Anonymous example, we have $Y = N$ and $X_0 = \mathbb{N}$, and all the assumptions hold. Clearly $Y$ is compact iff $X$ is locally compact.

Proof:

Suppose $K$ is a compactification of $X$ and let $S\subseteq K$ be a compact connected set with more than two points. Note that $\beta X = \beta X_0$ since $X_0\subseteq X \subseteq \beta X_0$ and so there exists a surjective map $\beta X_0\to K$ which maps $X$ to itself and $Y$ to $K\setminus X$.

Since $\beta X_0\setminus \text{cl}_{\beta X_0} Y \subseteq K$ is open, given $p\in S \cap (\beta X_0\setminus \text{cl}_{\beta X_0} Y)$ and $q\in S, q\neq p$, there exists a clopen compact subset $U$ of $\beta X_0\setminus \text{cl}_{\beta X_0} Y$ with $p\in U, q\notin U$, which is impossible. So that we need to have $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y\cup K\setminus X$. Let $L = \text{cl}_{\beta X_0} Y \setminus Y\cup K\setminus X$, $L$ compact.

Since $Y$ is locally compact, $\text{cl}_{\beta X_0} Y \setminus Y$ is closed in $L$, so $K\setminus X$ is open in $L$. The map $Y\to K\setminus X$ is perfect and so by this answer by K. P. Hart, $K\setminus X$ is open and zero-dimensional. If now $p\in S\cap (K\setminus X)$ and $q\in S, p\neq q$, then we can take a clopen compact subset $U$ of $L$ with $p\in U, q\notin U$, so as previously we can assume $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$.

Since $\text{cl}_{\beta X_0} Y \setminus Y$ is subspace of zero-dimensional space, it's itself zero-dimensional. And so there can't be a connected $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$ with more than two points.

It follows that $K$ is totally disconnected and compact, and so zero-dimensional. $\square$

Example. If $Y$ is a strongly zero-dimensional, locally compact, scattered space, then there exists a space $X$ with only zero-dimensional compactifications and $\beta X\setminus X\cong Y$.

Proof: Let $\alpha > 0$ be a non-limit ordinal with $|\beta Y| < \aleph_\alpha$ and $X_0 = \beta Y\times \omega_\alpha$. Then $\beta X_0 = \beta Y \times (\omega_\alpha + 1)$ is zero-dimensional. The space $X = \beta X_0\setminus (Y\times \{\omega_\alpha\})$ has only zero-dimensional compactifications and is such that $\beta X\setminus X \cong Y$. $\square$

Corollary. Let $X$ be a strongly zero-dimensional Tychonoff space with $\beta X\setminus X$ locally compact. Then the following are equivalent:

1. $X$ has only zero-dimensional compactifications.

2. $\beta X\setminus X$ is scattered.

Proof: 2 implies 1 from the theorem.

Suppose $X$ has only zero-dimensional compactifications. Then $\beta X\setminus X$ is zero-dimensional as a subspace of $\beta X$. If $\beta X\setminus X$ weren't scattered, then we would find a compact perfect set $A\subseteq \beta X\setminus X$ as in K. P. Hart answer and so there would exist a surjection $A\to [0, 1]$. Since $\beta X\setminus A$ is locally compact, from Engelking theorem 3.5.13 there exists a compactification $Z$ of $\beta X\setminus A$ with remainder homeomorphic to $[0, 1]$. Its also a compactification of $X$, and $Z$ is not zero-dimensional. The contradiction shows that such set $A$ cannot exist, and so $\beta X\setminus X$ is scattered. $\square$

Theorem. Let $X_0$ be a strongly zero-dimensional Tychonoff space, $Y\subseteq \beta X_0\setminus X_0$ is $C^*$-embedded subspace, $Y$ non-compact and locally compact, strongly zero-dimensional, scattered space. Then $X = \beta X_0\setminus Y$ is not locally compact, $\beta X\setminus X = Y$ and $X$ has only zero-dimensional compactifications.

Remark. In Anonymous example, we have $Y = N$ and $X_0 = \mathbb{N}$, and all the assumptions hold.

Proof: Let $y\in \text{cl}_{\beta X_0} Y \setminus Y$ which is non-empty since $Y$ isn't compact, then if $U$ is a compact clopen neighbourhood of $y$ in $X$ and $y\in V\cap X\subseteq U$ where $V\subseteq \beta X_0$ is open, then since $X$ is dense in $\beta X_0$ we have $\overline{V\cap X} = \overline{V}$ and so $V\subseteq U$, which is impossible since $V\cap Y\neq \emptyset$ but $U\cap Y = \emptyset$.

Suppose $K$ is a compactification of $X$ and let $S\subseteq K$ be a compact connected set with more than two points. Note that $\beta X = \beta X_0$ since $X_0\subseteq X \subseteq \beta X_0$ and so there exists a surjective map $\beta X_0\to K$ which maps $X$ to itself and $Y$ to $K\setminus X$.

Since $Y$ is strongly zero-dimensional, $\beta Y$ is zero-dimensional, and so $\beta Y\setminus Y$ is zero-dimensional. It follows that because $Y$ is $C^*$-embedded in $\beta X_0$, $\text{cl}_{\beta X_0} Y \setminus Y \cong \beta Y\setminus Y$ is zero-dimensional. And so there can't be a connected $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$ with more than two points.

Example. If $Y$ is a non-compact, locally compact, strongly zero-dimensional, scattered space, then there exists a space $X$ with only zero-dimensional compactifications and $\beta X\setminus X\cong Y$.

Proof: Let $\alpha > 0$ be a non-limit ordinal with $|\beta Y| < \aleph_\alpha$ and $X_0 = \beta Y\times \omega_\alpha$. Then $\beta X_0 = \beta Y \times (\omega_\alpha + 1)$ and $Y\times \{\omega_\alpha\}$ is $C^*$-embedded in $\beta X_0$, since given a bounded continuous function $Y\times \{\omega_\alpha\}\to \mathbb{R}$ we can extend it to $\beta Y\times \{\omega_\alpha\}\to \mathbb{R}$ and then to whole of $\beta X_0$ since $\beta Y\times \{\omega_\alpha\}$ is compact. The space $X = \beta X_0\setminus (Y\times \{\omega_\alpha\})$ from above theorem is such that $\beta X = \beta X_0$ and so it has remainder $\beta X\setminus X \cong Y$. $\square$

Theorem. Let $X_0$ be a strongly zero-dimensional Tychonoff space, $Y\subseteq \beta X_0\setminus X_0$ a locally compact scattered space. Then $X = \beta X_0\setminus Y$ is such that $\beta X\setminus X = Y$ and $X$ has only zero-dimensional compactifications.

Remark. In Anonymous example, we have $Y = N$ and $X_0 = \mathbb{N}$, and all the assumptions hold. Clearly $Y$ is compact iff $X$ is locally compact.

Proof:

Suppose $K$ is a compactification of $X$ and let $S\subseteq K$ be a compact connected set with more than two points. Note that $\beta X = \beta X_0$ since $X_0\subseteq X \subseteq \beta X_0$ and so there exists a surjective map $\beta X_0\to K$ which maps $X$ to itself and $Y$ to $K\setminus X$.

Since $\text{cl}_{\beta X_0} Y \setminus Y$ is subspace of zero-dimensional space, it's itself zero-dimensional. And so there can't be a connected $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$ with more than two points.

Example. If $Y$ is a locally compact, scattered space, then there exists a space $X$ with only zero-dimensional compactifications and $\beta X\setminus X\cong Y$.

Proof: Let $\alpha > 0$ be a non-limit ordinal with $|\beta Y| < \aleph_\alpha$ and $X_0 = \beta Y\times \omega_\alpha$. Then $\beta X_0 = \beta Y \times (\omega_\alpha + 1)$. The space $X = \beta X_0\setminus (Y\times \{\omega_\alpha\})$ is such that $\beta X\setminus X \cong Y$. $\square$

Corollary. Let $X$ be a strongly zero-dimensional Tychonoff space with $\beta X\setminus X$ locally compact. Then the following are equivalent:

1. $X$ has only zero-dimensional compactifications.

2. $\beta X\setminus X$ is scattered.

Proof: 2 implies 1 from the theorem.

Suppose $X$ has only zero-dimensional compactifications. Then $\beta X\setminus X$ is zero-dimensional as a subspace of $\beta X$. If $\beta X\setminus X$ weren't scattered, then we would find a compact perfect set $A\subseteq \beta X\setminus X$ as in K. P. Hart answer and so there would exist a surjection $A\to [0, 1]$. Since $\beta X\setminus A$ is locally compact, from Engelking theorem 3.5.13 there exists a compactification $Z$ of $\beta X\setminus A$ with remainder homeomorphic to $[0, 1]$. Its also a compactification of $X$, and $Z$ is not zero-dimensional. The contradiction shows that such set $A$ cannot exist, and so $\beta X\setminus X$ is scattered. $\square$

This is to finish argument of Anonymous that $X$ has only zero-dimensional compactifications.

Theorem. Let $X_0$ be a Tychonoff space, $Y\subseteq \beta X_0\setminus X_0$ is $C^*$-embedded subspace, $Y$ non-compact and locally compact, strongly zero-dimensional, scattered space. Then $X = \beta X_0\setminus Y$ is not locally compact, $\beta X\setminus X = Y$ and $X$ has only zero-dimensional compactifications.

Remark. In Anonymous example, we have $Y = N$ and $X_0 = \mathbb{N}$, and all the assumptions hold.

Proof: Let $y\in \text{cl}_{\beta X_0} Y \setminus Y$ which is non-empty since $Y$ isn't compact, then if $U$ is a compact clopen neighbourhood of $y$ in $X$ and $y\in V\cap X\subseteq U$ where $V\subseteq \beta X_0$ is open, then since $X$ is dense in $\beta X_0$ we have $\overline{V\cap X} = \overline{V}$ and so $V\subseteq U$, which is impossible since $V\cap Y\neq \emptyset$ but $U\cap Y = \emptyset$.

Suppose $K$ is a compactification of $X$ and let $S\subseteq K$ be a compact connected set with more than two points. Note that $\beta X = \beta X_0$ since $X_0\subseteq X \subseteq \beta X_0$ and so there exists a surjective map $\beta X_0\to K$ which maps $X$ to itself and $Y$ to $K\setminus X$.

Since $\beta X_0\setminus \text{cl}_{\beta X_0} Y \subseteq K$ is open, given $p\in S \cap (\beta X_0\setminus \text{cl}_{\beta X_0} Y)$ and $q\in S, q\neq p$, there exists a clopen compact subset $U$ of $\beta X_0\setminus \text{cl}_{\beta X_0} Y$ with $p\in U, q\notin U$, which is impossible. So that we need to have $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y\cup K\setminus X$. Let $L = \text{cl}_{\beta X_0} Y \setminus Y\cup K\setminus X$, $L$ compact.

Since $Y$ is locally compact, $\text{cl}_{\beta X_0} Y \setminus Y$ is closed in $L$, so $K\setminus X$ is open in $L$. The map $Y\to K\setminus X$ is perfect and so by this answer by K. P. Hart, $K\setminus X$ is open and zero-dimensional. If now $p\in S\cap (K\setminus X)$ and $q\in S, p\neq q$, then we can take a clopen compact subset $U$ of $L$ with $p\in U, q\notin U$, so as previously we can assume $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$.

Since $Y$ is strongly zero-dimensional, $\beta Y$ is zero-dimensional, and so $\beta Y\setminus Y$ is zero-dimensional. It follows that because $Y$ is $C^*$-embedded in $\beta X_0$, $\text{cl}_{\beta X_0} Y \setminus Y \cong \beta Y\setminus Y$ is zero-dimensional. And so there can't be a connected $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$ with more than two points.

It follows that $K$ is totally disconnected and compact, and so zero-dimensional. $\square$

Example. If $Y$ is a non-compact, locally compact, strongly zero-dimensional, scattered space, then there exists a space $X$ with only zero-dimensional compactifications and $\beta X\setminus X\cong Y$.

Proof: Let $\alpha > 0$ be a non-limit ordinal with $|\beta Y| < \aleph_\alpha$ and $X_0 = \beta Y\times \omega_\alpha$. Then $\beta X_0 = \beta Y \times (\omega_\alpha + 1)$ and $Y\times \{\omega_\alpha\}$ is $C^*$-embedded in $\beta X_0$, since given a bounded continuous function $Y\times \{\omega_\alpha\}\to \mathbb{R}$ we can extend it to $\beta Y\times \{\omega_\alpha\}\to \mathbb{R}$ and then to whole of $\beta X_0$ since $\beta Y\times \{\omega_\alpha\}$ is compact. The space $X = \beta X_0\setminus (Y\times \{\omega_\alpha\})$ from above theorem is such that $\beta X = \beta X_0$ and so it has remainder $\beta X\setminus X \cong Y$. $\square$

This is based on Anonymous answer.

Theorem. Let $X_0$ be a strongly zero-dimensional Tychonoff space, $Y\subseteq \beta X_0\setminus X_0$ is $C^*$-embedded subspace, $Y$ non-compact and locally compact, strongly zero-dimensional, scattered space. Then $X = \beta X_0\setminus Y$ is not locally compact, $\beta X\setminus X = Y$ and $X$ has only zero-dimensional compactifications.

Remark. In Anonymous example, we have $Y = N$ and $X_0 = \mathbb{N}$, and all the assumptions hold.

Proof: Let $y\in \text{cl}_{\beta X_0} Y \setminus Y$ which is non-empty since $Y$ isn't compact, then if $U$ is a compact clopen neighbourhood of $y$ in $X$ and $y\in V\cap X\subseteq U$ where $V\subseteq \beta X_0$ is open, then since $X$ is dense in $\beta X_0$ we have $\overline{V\cap X} = \overline{V}$ and so $V\subseteq U$, which is impossible since $V\cap Y\neq \emptyset$ but $U\cap Y = \emptyset$.

Suppose $K$ is a compactification of $X$ and let $S\subseteq K$ be a compact connected set with more than two points. Note that $\beta X = \beta X_0$ since $X_0\subseteq X \subseteq \beta X_0$ and so there exists a surjective map $\beta X_0\to K$ which maps $X$ to itself and $Y$ to $K\setminus X$.

Since $\beta X_0\setminus \text{cl}_{\beta X_0} Y \subseteq K$ is open, given $p\in S \cap (\beta X_0\setminus \text{cl}_{\beta X_0} Y)$ and $q\in S, q\neq p$, there exists a clopen compact subset $U$ of $\beta X_0\setminus \text{cl}_{\beta X_0} Y$ with $p\in U, q\notin U$, which is impossible. So that we need to have $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y\cup K\setminus X$. Let $L = \text{cl}_{\beta X_0} Y \setminus Y\cup K\setminus X$, $L$ compact.

Since $Y$ is locally compact, $\text{cl}_{\beta X_0} Y \setminus Y$ is closed in $L$, so $K\setminus X$ is open in $L$. The map $Y\to K\setminus X$ is perfect and so by this answer by K. P. Hart, $K\setminus X$ is open and zero-dimensional. If now $p\in S\cap (K\setminus X)$ and $q\in S, p\neq q$, then we can take a clopen compact subset $U$ of $L$ with $p\in U, q\notin U$, so as previously we can assume $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$.

Since $Y$ is strongly zero-dimensional, $\beta Y$ is zero-dimensional, and so $\beta Y\setminus Y$ is zero-dimensional. It follows that because $Y$ is $C^*$-embedded in $\beta X_0$, $\text{cl}_{\beta X_0} Y \setminus Y \cong \beta Y\setminus Y$ is zero-dimensional. And so there can't be a connected $S\subseteq \text{cl}_{\beta X_0} Y \setminus Y$ with more than two points.

It follows that $K$ is totally disconnected and compact, and so zero-dimensional. $\square$

Example. If $Y$ is a non-compact, locally compact, strongly zero-dimensional, scattered space, then there exists a space $X$ with only zero-dimensional compactifications and $\beta X\setminus X\cong Y$.

Proof: Let $\alpha > 0$ be a non-limit ordinal with $|\beta Y| < \aleph_\alpha$ and $X_0 = \beta Y\times \omega_\alpha$. Then $\beta X_0 = \beta Y \times (\omega_\alpha + 1)$ and $Y\times \{\omega_\alpha\}$ is $C^*$-embedded in $\beta X_0$, since given a bounded continuous function $Y\times \{\omega_\alpha\}\to \mathbb{R}$ we can extend it to $\beta Y\times \{\omega_\alpha\}\to \mathbb{R}$ and then to whole of $\beta X_0$ since $\beta Y\times \{\omega_\alpha\}$ is compact. The space $X = \beta X_0\setminus (Y\times \{\omega_\alpha\})$ from above theorem is such that $\beta X = \beta X_0$ and so it has remainder $\beta X\setminus X \cong Y$. $\square$