The answer to the question ``Does there exist $\dots$'' is Yes.

Let $X = [0,\infty)$ be the nonnegative part of the real line.
Let $\tau$ be the topology on $X$ consisting of all sets $[0,r)$
for $r\in X$. Let $\sigma$ be the indiscrete topology on $X$ and let
$\rho$ be the restriction of the usual topology of the real line to
$X$.

Claim 1. $[\sigma,\tau)$ has no maximal element.

Claim 2. $(\tau,\rho]$ has no minimal element.

Proof of Claim 1:
Given a topology $\tau'$ contained in $\tau$, let
$U(\tau') = \{r\in X\;|\; [0,r)\in\tau'\}$. That is,
$U(\tau')$ is the set of ``upper limits'' of open sets in $\tau'$.
Observe that if $\tau'$ is contained in $\tau$, then
$U(\tau')$ is a subset of $X=[0,\infty)$ that contains $0$
and is closed under the formation of suprema (since $\tau'$
is closed under the formation of unions).

Now choose and fix some $\tau'$ contained in $\tau$.

Case 1: $U(\tau')$ is dense in $X$.

In this case, since $U(\tau')$ is closed under sup's,
$U(\tau')=X$, so $\tau'=\tau$. Such a $\tau'$ is not
in $[\sigma,\tau)$.

Case 2: $U(\tau')$ is not dense in $X$.

There is a small interval
$(a-\epsilon,a+\epsilon)$ contained in $X$ and disjoint
from $U(\tau')$. The topology generated by $\tau'\cup\{[0,a)\}$
strictly extends $\tau'$ yet is still contained in $\tau$.
This shows that $\tau'$ is not maximal in this case.

This completes the proof of Claim 1.

Remember that Claim 2 asserts ``$(\tau,\rho]$ has no minimal element''.

Proof of Claim 2:
Let's reuse the notation $\tau'$, now for a proposed minimal element of $(\tau,\rho]$.
Necessarily $\tau'$ can be generated by $\tau$ and one set $A\in\rho\setminus\tau$.
Since $A$ is not in $\tau$, it is either (i) not connected in $\langle X;\rho\rangle$
or else (ii) does not
contain the element $0$. Whichever is the case, it is possible to find
$r\in X$ such that $B:=[0,r)\cap A\;(\in\rho)$ is a proper subset of
$A$ that is still not in $\tau$. To finish, we argue that the topology
generated by $\tau\cup\{B\}$ does not contain $A$, hence is strictly
smaller than $\tau'$.

The topology generated by $\tau\cup\{B\}$ consists of (i) sets from $\tau$,
(ii) sets of the form $V\cap B$ where $V\in\tau$, or (iii) any union of
a set from (i) with a set from (ii). $A$ is not of this form.
By construction, $\sup(B)<\sup(A)$. Any set $W$ of any of the three types
which does not belong to $\tau$ must have $\sup(W)\leq\sup(B)<\sup(A)$,
hence cannot equal $A$.