Find any positive integers $a,b,c,d,e$ so that $a,b^2,c^3,d^4,e^5$ form a nonconstant arithmetic progression.
Suppose integers $a,b,c,d,e$ are such that $a,b^2,c^3,d^4,e^5$ forms a nonconstant arithmetic progression with common difference $h \neq 0.$ Then $2b^2 = c^3 + a, 2d^4 = c^3 + e^5, 2c^3 = d^4+b^2, 2c^3 = a+e^5\tag{1}.$ Obviously the nonconstant condition is crucial because otherwise we could just take $a=b=c=d=e=1.$ Observe that all elements in $\{a,c,e\}$ and $\{b,d\}$ have the same parity. Note that none of $b,c,d$ can equal 1, for otherwise the sequence, being nonconstant, would decrease from the previous term to the term equal to 1 and then the next term will be nonpositive. First try $a=1$. We need $2b^2=c^3 +1 = (c+1)(c^2 - c+1)$. If $c$ is even, then $c^3+1$ is odd, a contradiction. So $c=2c_1+1$ for some positive integer $c_1$. Then $2b^2 = (2c_1+2)((2c_1+1)^2 - (2c_1+1) + 1) = 2(c_1+1)(4c_1^2+2c_1+1) \Rightarrow $ So perhaps . Now suppose $e = 1$. Then the sequence is necessarily decreasing and we have from (1) that $c^3 = \dfrac{a+1}2, 2d^4 = c^3 + 1= \dfrac{a+3}2\Rightarrow d^4 = \dfrac{a+3}4, 2b^2 = \dfrac{a+1}2 +a\Rightarrow b^2 = \dfrac{3a+1}4$. And we also have from (1) that $a+1 = \dfrac{a+3}4 + \dfrac{3a+1}4.$ Trying $c=3$ gives $a=53, b^2 = 40 = 5\cdot 2^3, c=5\Rightarrow a=249 = 3\cdot 83, b^2=187, c=7\Rightarrow a=514=2\cdot 257, c=9\Rightarrow a = 1457, b^2=1093$. There doesn't seem to be any good pattern with the $b^2$'s produced that gives away a modular arithmetic argument.
I think a lot of small possibilities won't work; for instance $e = 3$ implies $d\ge 4,$ but $d=4,5,6$ all fail to work.
I don't see any particular easy way to finish using your approach. Instead, note the exponents are all relatively prime to each other, except for $2$ and $4$. Thus, if we consider scaling an initial set of integers in an arithmetic sequence so each value then has an appropriate multiple of prime factors to match the requirements, the parities of the exponents for each of the prime factor in $b^2$ and $d^4$ must match.
Thus, we can use $2(3^2) = 18$ and $2^5 = 32$ as those initial values (note this may not the simplest case, but my initial checks didn't find anything easier and it works well enough here), to then get a starting sequence, with a common difference of $7$, along with a specific multiplier $m$, of
$$\{11, 2(3^2), 5^2, 2^5, 3(13)\}, \; m = 2^{f}3^{g}5^{h}13^{i} \tag{1}\label{eq1A}$$
For each of these $5$ values in the set to be multiplied by $m$ to form your sequence, we require (not including $a$ since it's not a limiting issue)
$$b^2 = 2^{f+1}3^{g+2}5^{h}13^{i}, \; c^3 = 2^{f}3^{g}5^{h+2}13^{i}, \; d^4 = 2^{f+5}3^{g}5^{h}13^{i}, \; e^5 = 2^{f}3^{g+1}5^{h}13^{i+1} \tag{2}\label{eq2A}$$
Combining all of the required exponent modulo values will produce results in modulo $\operatorname{lcm}(1,2,3,4,5)=60$. Since the requirement modulo $2$ is subsumed by that of modulo $4$, we won't include checks for it here. Thus, first considering the exponent of $2$, we thus have
$$f \equiv 0 \pmod{3}, \; f \equiv 3 \pmod{4}, \; f \equiv 0 \pmod{5} \tag{3}\label{eq3A}$$
The first & third equation combine to show that $f \equiv 0 \pmod{15}$. Adding multiples of $15$ to get the second equation results in
$$f \equiv 15 \pmod{60} \tag{4}\label{eq4A}$$
meeting all of the criteria. Following this same procedure, for the exponent of $3$, we get
$$g \equiv 0 \pmod{3}, \; g \equiv 0 \pmod{4}, \; g \equiv 4 \pmod{5} \; \to \; g \equiv 24 \pmod{60} \tag{5}\label{eq5A}$$
Continuing with the exponent of $5$, we next get
$$h \equiv 1 \pmod{3}, \; h \equiv 0 \pmod{4}, \; h \equiv 0 \pmod{5} \; \to \; h \equiv 40 \pmod{60} \tag{6}\label{eq6A}$$
Finally, with the exponent of $13$, we have
$$i \equiv 0 \pmod{3}, \; i \equiv 0 \pmod{4}, \; i \equiv 4 \pmod{5} \; \to \; i \equiv 24 \pmod{60} \tag{7}\label{eq7A}$$
Thus, putting everything together, we have that for any integer $k \ge 0$, the set below
$$\{11m, 18m, 25m, 32m, 39m\}, \; m = 2^{60k+15}3^{60k+24}5^{60k+40}13^{60k+24} \tag{8}\label{eq8A}$$
forms an arithmetic sequence, with a common difference of $7m$, along with $a = 11m$, $b^2 = 2^{60k+16}3^{60k+26}5^{60k+40}13^{60k+24} \; \to \; b = 2^{30k+8}3^{30k+13}5^{30k+20}13^{30k+12}$, $c^3 = 2^{60k+15}3^{60k+24}5^{60k+42}13^{60k+24} \; \to \; c = 2^{20k+5}3^{20k+8}5^{20k+14}13^{20k+8}$, $d^4 = 2^{60k+20}3^{60k+24}5^{60k+40}13^{60k+24} \; \to \; d = 2^{15k+5}3^{15k+6}5^{15k+10}13^{15k+6}$ and $e^5 = 2^{60k+20}3^{60k+25}5^{60k+40}13^{60k+25} \; \to \; e = 2^{12k+4}3^{12k+5}5^{12k+8}13^{12k+5}$.
More generally, for any integer $j \gt 1$, multiplying each element of any valid arithmetic sequence by $j^{60}$ produces a new valid sequence.
Update: A somewhat simpler example (e.g., the multiplier only requires $3$ primes instead of $4$, the modulo equations are a bit easier to solve, and $1$ is one of the base terms) is $\{3(11), 5^2, 17, 3^2, 1\}$, giving for any integer $k \ge 0$ an $m = 3^{60k+30}5^{60k}17^{60k+20}$. Multiplying each term by $m$ gives an arithmetic sequence with a common difference of $-8m$, along with $a = 33m$, $b^2 = 3^{60k+30}5^{60k+2}17^{60k+20} \; \to \; b = 3^{30k+15}5^{30k+1}17^{30k+10}$, $c^3 = 3^{60k+30}5^{60k}17^{60k+21} \; \to \; c = 3^{20k+10}5^{20k}17^{20k+7}$, $d^4 = 3^{60k+32}5^{60k}17^{60k+20} \; \to \; d = 3^{15k+8}5^{15k}17^{15k+5}$ and $e^5 = 3^{60k+30}5^{60k}17^{60k+20} \; \to \; e = 3^{12k+6}5^{12k}17^{12k+4}$.