Update: So far I've been able to find (I believe all) solutions for $d=3,5,7$, and partially for $d=9$. If It turns out that a closed form for all solutions for a fixed (large enough) $d$ is too messy to compile, I'll probably include those as a partial self-answer here.
What is allowing me to solve and find closed forms for a fixed $d$ case now, is solving the related system with a CAS like Mathematica - and is now summarized in the answer below.
Question
Can we generate all numbers which are palindromic in two consecutive number bases $(b, b+1)$ and have $(2d+1, d\in\mathbb N)$ digits when written in their palindromic bases?
$3$ digit pattern
I've found the pattern for $d=1$, the $3$ digit palindromes rather straightforward (simple):
(Just look at the examples below and you'll see it)
All solutions that are palindromic in number bases $b\in\{n+1, n+2\}$ for $n\gt3$, are given by the following equation: (pick all $k$ in range to get a solution per value of $k$)
$$ P_3(n) = \begin{cases} n^3+2n^2+2n, & \text{$k=1$} \\ n^3-n^2(k-4)-n(k-7)-3k+4, & \text{$n\gt k \gt1$} \end{cases} $$
That gives you $n-1$ solutions, which are palindromic in $n+1,n+2$ bases.
The digits of given examples in base $b=n+1$ are: $[n-k+1], [(n-k+2) \mod (n+1)], [n-k+1]$
$\text{(Examples)}$ You can see the solutions for $n=4,5,6$ below:
5 67 3 [2, 3, 2] [1, 5, 1]
5 98 3 [3, 4, 3] [2, 4, 2]
5 104 3 [4, 0, 4] [2, 5, 2]
6 92 3 [2, 3, 2] [1, 6, 1]
6 135 3 [3, 4, 3] [2, 5, 2]
6 178 3 [4, 5, 4] [3, 4, 3]
6 185 3 [5, 0, 5] [3, 5, 3]
7 121 3 [2, 3, 2] [1, 7, 1]
7 178 3 [3, 4, 3] [2, 6, 2]
7 235 3 [4, 5, 4] [3, 5, 3]
7 292 3 [5, 6, 5] [4, 4, 4]
7 300 3 [6, 0, 6] [4, 5, 4]
Same pattern holds for all $n\gt3$, as observed.
First column is the base $b=n+1$, second is the number, third are the digits, and last two are number representations in number bases $(n+1,n+2)$.
The solutions for $n=2,3$ which are not included in $P_3(n)$ are $10, 46$ ;
Where number bases $(2,3)$, the $n=1$ case, has no solutions.
$5,7,9\dots$ digit pattern?
I've found patterns for $5,7,9,\dots$ digits to be similar among themselves but not so simple as the three digit one, as they seem to be more unpredictable with more digits.
Can we define a function / algorithm to generate all the solutions for some $P_d(n)$?
Meaning, the solutions are palindromic in bases $(n+1,n+2)$ and have $d$ digits?
$P_3(n)$ above is the simplest example, which generates all solutions for $(n+1,n+2)$ bases by iterating $k\in(n,1]$. How can one find patterns for other digit cases? (see linked code below to generate examples for some $2d+1$ digits)
Use linked code to generate solutions for any case of $2d+1$ digits by simply changing the digit = 4 variable to some other integer. (You can also generate from/to any number base)
$9$ digit pattern?
I'm interested in the $9$ digit pattern before the rest.
(Since this would help search for solutions for numbers palindromic in three consecutive bases which also have $9$ digits; where $3,5,7$ digit patterns were found, but no examples for $\ge9$ digits have been found. - I've posted these patterns in my question regarding numbers palindromic in four or more consecutive bases. The MSE version of the question is linked in the comments.)
You can see all solutions for bases up to $32$ here, which were generated by my python code.
Linked code checks every palindrome in some number base to see if it is palindromic in bases $(b, b+1)$; but this is extremely slow at larger bases, so I need to find $P_9(n)$ to generate them instead.
In the linked output text, you can notice patterns in digits. For example, notice patterns emerging at base $b=30$, for first couple examples; highlighted here.
Can you predict these patterns for some base $b$, at least to some degree? I couldn't.