To answer Q2: Every noetherian (commutative) ring that is not a finite product of fields admits irreducible elements. Since any finite product of fields admits prime elements, this shows that the answer to Q2 is no.

In an arbitrary commutative ring $A$, define $x\in A$ to be quasi-irreducible if $Ax$ is maximal among principal ideals $\neq A$. Thus if $A$ is noetherian and not a field, then it admits nonzero quasi-irreducible elements. (Note that 0 is quasi-irreducible iff $A$ is a field.) Moreover, irreducible implies quasi-irreducible.

The converse to holds for nonzero elements when $A$ is a domain, but also more generally when $A$ is connected, in the sense that it admits no nontrivial idempotent: (*) if $A$ is connected and $x\in A$ with $x\neq 0$, $x$ quasi-irreducible, then $x$ is irreducible.

Indeed, suppose that $x\neq 0$ is quasi-irreducible. Suppose by contradiction that $x=yz$ with $y,z$ non-invertible. Since $x$ is quasi-irreducible, this implies that $y,z\in Ax$. Write $y=ax$ and $z=bx$. Hence $x=abx^2$. That is, $x(1-abx)=0$. Then $(abx)^2=ab(abx^2)=abx$. Thus $abx$ is idempotent. Since $A$ is connected, either this implies $abx=1$, so $x$ is invertible (contradiction), or $abx=0$, whence $abx^2=x=0$, another contradiction.

Note that the converse to (*) does not hold in the non-connected case: for instance in a product of 2 fields, $(1,0)$ is quasi-irreducible but not irreducible.

Let us now prove the main claim. Assume that $A\neq 0$ is an arbitrary noetherian ring. Then it can be written as a product $A=A_1\times \dots \times A_k$ with each $A_i$ connected.

Suppose that one of the $A_i$ is not a field. After reindexing, we can suppose $i=1$. Let $x_1$ be a quasi-irreducible element in $A_1$; by (*) it is irreducible in $A_1$. Then $(x_1,1,\dots,1)$ is an irreducible element in $A$.

Finally, if all $A_i$ are fields, then $(0,1,\dots,1)$ is a prime element.

To answer Q2: Every noetherian (commutative) ring that is not a finite product of fields admits irreducible elements. Since any finite product of $\ge 2$ fields admits prime elements, this shows that the answer to Q2 is no.

In an arbitrary commutative ring $A$, define $x\in A$ to be quasi-irreducible if $Ax$ is maximal among principal ideals $\neq A$. Thus if $A$ is noetherian and not a field, then it admits nonzero quasi-irreducible elements. (Note that 0 is quasi-irreducible iff $A$ is a field.) Moreover, irreducible implies quasi-irreducible.

The converse to holds for nonzero elements when $A$ is a domain, but also more generally when $A$ is connected, in the sense that it admits no nontrivial idempotent: (*) if $A$ is connected and $x\in A$ with $x\neq 0$, $x$ quasi-irreducible, then $x$ is irreducible.

Indeed, suppose that $x\neq 0$ is quasi-irreducible. Suppose by contradiction that $x=yz$ with $y,z$ non-invertible. Since $x$ is quasi-irreducible, this implies that $y,z\in Ax$. Write $y=ax$ and $z=bx$. Hence $x=abx^2$. That is, $x(1-abx)=0$. Then $(abx)^2=ab(abx^2)=abx$. Thus $abx$ is idempotent. Since $A$ is connected, either this implies $abx=1$, so $x$ is invertible (contradiction), or $abx=0$, whence $abx^2=x=0$, another contradiction.

Note that the converse to (*) does not hold in the non-connected case: for instance in a product of 2 fields, $(1,0)$ is quasi-irreducible but not irreducible.

Let us now prove the main claim. Assume that $A\neq 0$ is an arbitrary noetherian ring. Then it can be written as a product $A=A_1\times \dots \times A_k$ with each $A_i$ connected.

Suppose that one of the $A_i$ is not a field. After reindexing, we can suppose $i=1$. Let $x_1$ be a quasi-irreducible element in $A_1$; by (*) it is irreducible in $A_1$. Then $(x_1,1,\dots,1)$ is an irreducible element in $A$.

Finally, if all $A_i$ are fields and $k\ge 2$, then $(0,1,\dots,1)$ is a prime element.

To answer Q2: Every noetherian domain that is not a finite product of fields admits irreducible elements. Since any finite product of fields admits prime elements, this shows that the answer to Q2 is no.

In an arbitrary commutative ring $A$, define $x\in A$ to be quasi-irreducible if $Ax$ is maximal among principal ideals $\neq A$. Thus if $A$ is noetherian and not a field, then it admits nonzero quasi-irreducible elements. (Note that 0 is quasi-irreducible iff $A$ is a field.) Moreover, irreducible implies quasi-irreducible.

The converse to holds for nonzero elements when $A$ is a domain, but also more generally when $A$ is connected, in the sense that it admits no nontrivial idempotent: (*) if $A$ is connected and $x\in A$ with $x\neq 0$, $x$ quasi-irreducible, then $x$ is irreducible.

Indeed, suppose that $x\neq 0$ is quasi-irreducible. Suppose by contradiction that $x=yz$ with $y,z$ non-invertible. Since $x$ is quasi-irreducible, this implies that $y,z\in Ax$. Write $y=ax$ and $z=bx$. Hence $x=abx^2$. That is, $x(1-abx)=0$. Then $(abx)^2=ab(abx^2)=abx$. Thus $abx$ is idempotent. Since $A$ is connected, either this implies $abx=1$, so $x$ is invertible (contradiction), or $abx=0$, whence $abx^2=x=0$, another contradiction.

Note that the converse to (*) does not hold in the non-connected case: for instance in a product of 2 fields, $(1,0)$ is quasi-irreducible but not irreducible.

Let us now prove the main claim. Assume that $A\neq 0$ is an arbitrary noetherian domain. Then it can be written as a product $A=A_1\times \dots \times A_k$ with each $A_i$ connected.

Suppose that one of the $A_i$ is not a field. After reindexing, we can suppose $i=1$. Let $x_1$ be a quasi-irreducible element in $A_1$; by (*) it is irreducible in $A_1$. Then $(x_1,1,\dots,1)$ is an irreducible element in $A$.

Finally, if all $A_i$ are fields, then $(0,1,\dots,1)$ is a prime element.

To answer Q2: Every noetherian (commutative) ring that is not a finite product of fields admits irreducible elements. Since any finite product of fields admits prime elements, this shows that the answer to Q2 is no.

In an arbitrary commutative ring $A$, define $x\in A$ to be quasi-irreducible if $Ax$ is maximal among principal ideals $\neq A$. Thus if $A$ is noetherian and not a field, then it admits nonzero quasi-irreducible elements. (Note that 0 is quasi-irreducible iff $A$ is a field.) Moreover, irreducible implies quasi-irreducible.

The converse to holds for nonzero elements when $A$ is a domain, but also more generally when $A$ is connected, in the sense that it admits no nontrivial idempotent: (*) if $A$ is connected and $x\in A$ with $x\neq 0$, $x$ quasi-irreducible, then $x$ is irreducible.

Indeed, suppose that $x\neq 0$ is quasi-irreducible. Suppose by contradiction that $x=yz$ with $y,z$ non-invertible. Since $x$ is quasi-irreducible, this implies that $y,z\in Ax$. Write $y=ax$ and $z=bx$. Hence $x=abx^2$. That is, $x(1-abx)=0$. Then $(abx)^2=ab(abx^2)=abx$. Thus $abx$ is idempotent. Since $A$ is connected, either this implies $abx=1$, so $x$ is invertible (contradiction), or $abx=0$, whence $abx^2=x=0$, another contradiction.

Note that the converse to (*) does not hold in the non-connected case: for instance in a product of 2 fields, $(1,0)$ is quasi-irreducible but not irreducible.

Let us now prove the main claim. Assume that $A\neq 0$ is an arbitrary noetherian ring. Then it can be written as a product $A=A_1\times \dots \times A_k$ with each $A_i$ connected.

Suppose that one of the $A_i$ is not a field. After reindexing, we can suppose $i=1$. Let $x_1$ be a quasi-irreducible element in $A_1$; by (*) it is irreducible in $A_1$. Then $(x_1,1,\dots,1)$ is an irreducible element in $A$.

Finally, if all $A_i$ are fields, then $(0,1,\dots,1)$ is a prime element.