Can it be shown, $n^4+(n+d)^4+(n+2d)^4\ne z^4$? We know, $n^4+(n+d)^4= z^4$ has no solution in positive integers $n,d,z$.
Can it be shown, $n^4+(n+d)^4+(n+2d)^4= z^4$ has no solution in positive integers $n,d,z$?
I am check upto $1\le n, d, z\le 150$ without finding a counter example.
PARI/GP
for(n=1,150,for(d=1,150,for(p=1,150,if(sum(q=0,2,(n+q*d)^4)==p^4,print([n,d,p])))))

Generalization over problem
 A: First, assume there's at least one solution and then let $m = n + d$, with $m$ being the smallest positive integer which works. The equation then becomes
$$\begin{equation}\begin{aligned}
z^4 & = (m - d)^4 + m^4 + (m + d)^4 \\
& = m^4 - 4md^3 + 6m^2d^2 - 4m^3d + d^4 + m^4 + \\
& \; \; \; \; m^4 + 4md^3 + 6m^2d^2 + 4m^3d + d^4 \\
& = 3m^4 + 12m^2d^2 + 2d^4
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Next, note $a^4 \equiv 1 \pmod 3$ for all integers $a$ which aren't a multiple of $3$. The right side of \eqref{eq1A} is congruent to $2d^4$ modulo $3$, i.e., $2$ if $d$ is not a multiple of $3$, which is not possible so $d$ must be a multiple of $3$.
Thus, the right side of \eqref{eq1A} has a factor $3$, so $z$ is also a multiple of $3$, which means the left side has at least $4$ factors of $3$. If $m$ is not a multiple of $3$, then the right side has only $1$ factor of $3$ (since $12m^2d^2 + 2d^4$ has at least $3$ factors of $3$), which is not possible. Thus, $m$ must also be a multiple of $3$. This means we can divide both sides of \eqref{eq1A} by $3^4 = 81$, which is equivalent to dividing $z$, $m$ and $d$ each by $3$ to, say, the $3$ integers $z_1 = \frac{z}{3}$, $m_1 = \frac{m}{3}$ and $d_1 = \frac{d}{3}$, which then results in
$$z_1^4 = (m_1 - d_1)^4 + m_1^4 + (m_1 + d_1)^4 \tag{2}\label{eq2A}$$
Note this is of the same form as \eqref{eq1A}, but with $m_1$ being a smaller positive integer than $m$ which works. However, initially, I set $m$ to be the smallest such integer. This is a contradiction, so there's no smallest positive value for $m$ and, thus, no solution at all.
