Regarding Question 1:

It is certainly possible (i.e., consistent with ZFC) that there is only one transitive model of ZFC. For example, if there are transitive models $M$ and $N$ of ZFC such that $o(M)<o(N)$ (where $o(M)$ is the ordinal height of $M$), then there are ordinals $\alpha < \beta$ such that $L(\alpha)$ is the so-called Shepherdson-Cohen minimal model of set thery, which is the smallest transitive model of ZFC (where $L(\alpha)$ is the $\alpha$-th approximation to the constructible universe), and $L(\beta)$ is the next shortest transitive model of ZFC. In this situation $L(\beta)$ is a model of ZFC in which there is no transitive model of ZFC of height greater than $\alpha$.

So this shows that there is no universal technique within ZFC for making a transitive model of ZFC taller (where "$N$ is taller than $M$" is defined as $o(M)<o(N)$).

Regarding Question 2.

This question has already been answered by Carl Mummert's comment if "taller" is defined via: $N$ is taller than $M$ iff $o(M)<o(N)$ [since in the theory ZFC plus there are arbitrarily large inaccessibles, every transitive model of ZFC is shorter than some other transitive model].

On the other hand, if "taller'' is defined by the current version of the question which stipulates that $N$ is taller than $M$ if there is an ordinal $\alpha \in N$ such that $M = V_{\alpha}^{N}$, then the answer is in the negative since the minimal model of set theory $L(\alpha)$ (mentioned the answer above to Question 1) has no such taller extension. To see this, one needs to use the well-known fact that $L(\alpha)$ is pointwise definable [I will try to provide a reference in my next edit]. If $N$ is a model of ZFC such that $M = V_{\alpha}^{N}$, then $N$ would have to "notice" that its own $V_{\alpha}$ is countable (since $N$ can define define the satisfaction predicate for any set structure, a satisfaction predicate that will agree with the "real world" satisfaction predicate for $L(\alpha)$), which contradicts an easy theorem of ZFC that says that if $\alpha > \omega$ then $V_{\alpha}$ is uncountable.

Regarding Question 1:

It is certainly possible (i.e., consistent with ZFC) that there is only one transitive model of ZFC. For example, if there are transitive models $M$ and $N$ of ZFC such that $o(M)<o(N)$ (where $o(M)$ is the ordinal height of $M$), then there are ordinals $\alpha < \beta$ such that $L(\alpha)$ is the so-called Shepherdson-Cohen minimal model of set thery, which is the smallest transitive model of ZFC (where $L(\alpha)$ is the $\alpha$-th approximation to the constructible universe), and $L(\beta)$ is the next shortest transitive model of ZFC + V=L. In this situation $L(\beta)$ is a model of ZFC in which there is no transitive model of ZFC of height greater than $\alpha$.

So this shows that there is no universal technique within ZFC for making a transitive model of ZFC taller (where "$N$ is taller than $M$" is defined as $o(M)<o(N)$).

Regarding Question 2.

This question has already been answered by Carl Mummert's comment if "taller" is defined via: $N$ is taller than $M$ iff $o(M)<o(N)$ [since in the theory ZFC plus there are arbitrarily large inaccessibles, every transitive model of ZFC is shorter than some other transitive model].

On the other hand, if "taller'' is defined by the current version of the question which stipulates that $N$ is taller than $M$ if there is an ordinal $\alpha \in N$ such that $M = V_{\alpha}^{N}$, then the answer is in the negative since the minimal model of set theory $L(\alpha)$ (mentioned the answer above to Question 1) has no such taller extension. To see this, one needs to use the well-known fact that $L(\alpha)$ is pointwise definable [I will try to provide a reference in my next edit]. If $N$ is a model of ZFC such that $M = V_{\alpha}^{N}$, then $N$ would have to "notice" that its own $V_{\alpha}$ is countable (since $N$ can define define the satisfaction predicate for any set structure, a satisfaction predicate that will agree with the "real world" satisfaction predicate for $L(\alpha)$), which contradicts an easy theorem of ZFC that says that if $\alpha > \omega$ then $V_{\alpha}$ is uncountable.

Regarding Question 1:

It is certainly possible (i.e., consistent with ZFC) that there is only one transitive model of ZFC. For example, if there are transitive models $M$ and $N$ of ZFC such that $o(M)<o(N)$ (where $o(M)$ is the ordinal height of $M$), then there are ordinals $\alpha < \beta$ such that $L(\alpha)$ is the so-called Shepherdson-Cohen minimal model of set thery, which is the smallest transitive model of ZFC (where $L(\alpha)$ is the $\alpha$-th approximation to the constructible universe), and $L(\beta)$ is the next shortest transitive model of ZFC. In this situation $L(\beta)$ is a model of ZFC in which there is only one transitive model of ZFC, namely $L(\alpha$).

So this shows that there is no universal technique within ZFC for making a transitive model of ZFC taller (where "$N$ is taller than $M$ is defined as $o(M)<o(N)$).

Regarding Question 2.

This question has already been answered by Carl Mummert's comment if "taller" is defined via: $N$ is taller than $M$ iff $o(M)<o(N)$ [since in the theory ZFC plus there are arbitrarily large inaccessibles, every transitive model of ZFC is shorter than some other transitive model].

On the other hand, if "taller'' is defined by the current version of the question which stipulates that $N$ is taller than $M$ if there is an ordinal $\alpha \in N$ such that $M = V_{\alpha}^{N}$, then the answer is in the negative since the minimal model of set theory $L(\alpha)$ (mentioned the answer above to Question 1) has no such taller extension. To see this, one needs to use the well-known fact that $L(\alpha)$ is pointwise definable [I will try to provide a reference in my next edit]. If $N$ is a model of ZFC such that $M = V_{\alpha}^{N}$, then $N$ would have to "notice" that its own $V_{\alpha}$ is countable (since $N$ can define define the satisfaction predicate for any set structure, a satisfaction predicate that will agree with the "real world" satisfaction predicate for $L(\alpha)$), which contradicts an easy theorem of ZFC that says that if $\alpha > \omega$ then $V_{\alpha}$ is uncountable.

Regarding Question 1:

It is certainly possible (i.e., consistent with ZFC) that there is only one transitive model of ZFC. For example, if there are transitive models $M$ and $N$ of ZFC such that $o(M)<o(N)$ (where $o(M)$ is the ordinal height of $M$), then there are ordinals $\alpha < \beta$ such that $L(\alpha)$ is the so-called Shepherdson-Cohen minimal model of set thery, which is the smallest transitive model of ZFC (where $L(\alpha)$ is the $\alpha$-th approximation to the constructible universe), and $L(\beta)$ is the next shortest transitive model of ZFC. In this situation $L(\beta)$ is a model of ZFC in which there is no transitive model of ZFC of height greater than $\alpha$.

So this shows that there is no universal technique within ZFC for making a transitive model of ZFC taller (where "$N$ is taller than $M$" is defined as $o(M)<o(N)$).

Regarding Question 2.

This question has already been answered by Carl Mummert's comment if "taller" is defined via: $N$ is taller than $M$ iff $o(M)<o(N)$ [since in the theory ZFC plus there are arbitrarily large inaccessibles, every transitive model of ZFC is shorter than some other transitive model].

On the other hand, if "taller'' is defined by the current version of the question which stipulates that $N$ is taller than $M$ if there is an ordinal $\alpha \in N$ such that $M = V_{\alpha}^{N}$, then the answer is in the negative since the minimal model of set theory $L(\alpha)$ (mentioned the answer above to Question 1) has no such taller extension. To see this, one needs to use the well-known fact that $L(\alpha)$ is pointwise definable [I will try to provide a reference in my next edit]. If $N$ is a model of ZFC such that $M = V_{\alpha}^{N}$, then $N$ would have to "notice" that its own $V_{\alpha}$ is countable (since $N$ can define define the satisfaction predicate for any set structure, a satisfaction predicate that will agree with the "real world" satisfaction predicate for $L(\alpha)$), which contradicts an easy theorem of ZFC that says that if $\alpha > \omega$ then $V_{\alpha}$ is uncountable.