Let $p=2^n-1$ be a Mersenne prime. We know that $H=PSL(2,p^4)$ has an irreducible character of degree $ p^4$.


Is there any solvable group of order $H$ with an irreducible character of degree $ p^4$. This is equivalent to: if we consider a solvable group $G$ of order $p^4 (p+1)(p^2+1)(p^4+1)$, can we say that $ G $ has no irreducible complex character of degree $ p^4$?



Existence of some character degree in a solvable group
Given that $p$ is a Mersenne prime, it is easy to check that ${\rm gcd}(p+1,p^{2}+1) = {\rm gcd}(p+1,p^{4}+1)= {\rm gcd}(p^{2}+1,p^{4}+1) = 2$, and we recall that $p+1$ is a power of $2$. Let $G$ be a solvable group of order $p^{4}(p+1)(p^{2}+1)(p^{4}+1)$. Suppose that $G$ has an irreducible character of degree $p^{4}$. Then by results (eg of Brauer), we know that $O\_{p}(G) = 1$. Let us examine the structure of $O\_{p^{\prime}}(G)$.


Since $O\_{p}(G) = 1$, by the well-known centralizer Lemma of Hall-Higman, no element of order $p$ in $G$ acts trivially on $O\_{p^{\prime}}(G)$, so that $O\_{p^{\prime}}(G)$ admits a group of (outer) automorphisms $P$ of order $p^{4}$.


We claim though that every $P$-invariant section $M$ of $O\_{p^{\prime}}(G)$ of odd order is centralized by $P$. By standard results on coprime automorphisms, it suffices to assume that $M$ is an elementary $q$-group, and is minimal $P$-invariant, but not centralized by $P$. Then $|M|$ divides one (and only one) of $\frac{p^{2}+1}{2}$ or $\frac{p^{4}+1}{2}$, and $|M| \equiv 1$ (mod $p$) ( note for the purposes of dealing with other odd primes that for any odd prime $p$, $\frac{p+1}{2}$ clearly has no divisor (greater than $1$) which is congruent to $1$ (mod $p$)).


Suppose that $|M|$ divides $p^{2}+1$. Then $\frac{p^{2}+1}{|M|} \equiv 1$ (mod $p$), while $|M| \neq p^{2}+1$, since $p$ is Mersenne. Then we have the contradiction $p^{2}+1 = |M|\frac{p^{2}+1}{|M|} \geq (p+1)^{2}.$ Hence $|M|$ divides $p^{4}+1$.


By Y. Cor's answer to your previous question MO208645, we know that no divisor of $p^{4}+1$ other than $1$ and $p^{4}+1$ is congruent to $1$ (mod $p$). But we can't have $|M| = p^{4}+1$ since $|M|$ is a prime power and $p^{4}+1 \equiv 2$ (Mod $8$). This is a contradiction.


We conclude that for $P \in {\rm Syl}\_{p}(G)$, it must be the case that $[O\_{p^{\prime}}(G),P]$ is a $2$-group, say $X$. By standard results on coprime automorphisms, $[P,X] = X$ and $P$ acts faithfully on $X$.


Now a Sylow $2$-subgroup of $G$ has order $2^{n+2}$. Now $n$ is the smallest positive integer with $2^{n} \equiv 1$ (mod $p$). Now let $L = PX$. Then $|L|$ divides $2^{n+2}p^{4}$ and $[L:N\_{L}(P)]$ is a power of $2$, and is congruent to $1$ (mod $p$), but is greater than $1$.


It follows that $[L:N\_{L}(P)] = 2^{n}$ if $p > 3$, and that 
$[L:N\_{L}(P)] \in \{2^{n},2^{n+2} \}$ if $p = 3$. But then it is clear that $X$ admits no group of automorphisms of order $p^{4}$, which is a contradiction.


An alternative way to conclude is to take a minimal $P$-invariant $M$ section of $X$with $[P,M] \neq 1$. Then we have $|M| \geq 2^{n}$, and for every choice of $p$ (even $p=3$), there can be at most two such sections of $X$, each of which admits an automorphism of order $p$ but no group of automorphisms of order $p^{2}$. Hence in no case can $X$ admit a faiithful group of automorphisms of order $p^{4}$.


What has actually been proved is that when $p$ is a Mersenne prime, a $p$-solvable finite group $G$ of order $p^{4}(p+1)(p^{2}+1)(p^{4}+1)$ has a non-identity normal $p$-subgroup, so certainly has no $p$-block of defect zero.


( I don't think that the fact that $p$ is a Mersenne prime is essential here, I think the argument can be modified to work for any odd prime $p$. In fact, I believe the argument really shows that if $p$ is an odd prime, then a $p$-solvable group of order $p^{k}(p+1)(p^{2}+1)(p^{4}+1)$ has a non-identity normal $p$-subgroup if $k >1$ when $p \neq 3$ and if $k >2$ when $p =3$).


