In the cited paper by Wells Johnson, where the Theorem (there is only one, so it is unnumbered) sets $a^r = \pm 1 + tp$, instead set $a^r = a + 1 + p$, but maintain the underlying logic of the proof. Then this adaptation of his result gives $q_{a} = \frac{1}{r}(q_{a+1} - \frac{1}{a+1})$, which in your notation, assuming $p = a^n - a - 1$, is

$(a+1)nq_{p}(a) + 1 \equiv (a+1)q_{p}(a+1)$ (mod $p$),

the desired expression.

Your result, a consequence of Johnson's work that seems to have been previously overlooked, is interesting because it clearly reveals that for $p$ of the required forms, $q_{p}(2)$ and $q_{p}(3)$ cannot vanish simultaneously (a question addressed in the cited paper by Emma Lehmer).

In this vein, if instead of $a^r = a + 1 + p$ we take $a^r = a - 1 + p$, then $q_{a} = \frac{1}{r}(q_{a-1} - \frac{1}{a-1})$, or in your notation, assuming $p = a^n - a + 1$,

$(a-1)nq_{p}(a) + 1 \equiv (a-1)q_{p}(a-1)$ (mod $p$).

Thus, taking $a = 3$, if $p$ can be expressed in the form $3^n - 2$, we have

$2nq_{p}(3) + 1 \equiv 2q_{p}(2)$ (mod $p$).

Taking $a = 4$, if $p$ can be expressed in the form $4^n - 3$, we have

$3nq_{p}(4) + 1 \equiv 3q_{p}(3) \rightarrow 6nq_{p}(2) + 1 \equiv 3q_{p}(3)$ (mod $p$).

Taking $a = 9$, if $p$ can be expressed in the form $9^n - 8$, we have

$8nq_{p}(9) + 1 \equiv 8q_{p}(8) \rightarrow 16nq_{p}(3) + 1 \equiv 24q_{p}(2)$ (mod $p$).

The cases which relate $q_{p}(2)$ to $q_{p}(3)$ are merely some of the more interesting implications of the reformulation of Johnson's theorem suggested by your posting. Finally, if $p$ has the form $a^n -k$, where neither $a$ nor $k$ is divisible by $p$ nor greater than 89, a similar argument establishes the first case of Fermat's Last Theorem for $p$, since the failure thereof would require the vanishing of $q_{p}(a)$ for every $a \le 89$, as proved by Granville.

In the cited paper by Wells Johnson, where the Theorem sets $a^r = \pm 1 + tp$, instead set $a^r = a + 1 + p$, but maintain the underlying logic of the proof. Then this adaptation of his result gives $q_{a} = \frac{1}{r}(q_{a+1} - \frac{1}{a+1})$, which in your notation, assuming $p = a^n - a - 1$, is

$(a+1)nq_{p}(a) + 1 \equiv (a+1)q_{p}(a+1)$ (mod $p$),

the desired expression.

Your result, a consequence of Johnson's work that seems to have been previously overlooked, is interesting because it clearly reveals that for $p$ of the required forms, $q_{p}(2)$ and $q_{p}(3)$ cannot vanish simultaneously (a question addressed in the cited paper by Emma Lehmer).

In this vein, if instead of $a^r = a + 1 + p$ we take $a^r = a - 1 + p$, then $q_{a} = \frac{1}{r}(q_{a-1} - \frac{1}{a-1})$, or in your notation, assuming $p = a^n - a + 1$,

$(a-1)nq_{p}(a) + 1 \equiv (a-1)q_{p}(a-1)$ (mod $p$).

Thus, taking $a = 3$, if $p$ can be expressed in the form $3^n - 2$, we have

$2nq_{p}(3) + 1 \equiv 2q_{p}(2)$ (mod $p$).

Taking $a = 4$, if $p$ can be expressed in the form $4^n - 3$, we have

$3nq_{p}(4) + 1 \equiv 3q_{p}(3) \rightarrow 6nq_{p}(2) + 1 \equiv 3q_{p}(3)$ (mod $p$).

Taking $a = 9$, if $p$ can be expressed in the form $9^n - 8$, we have

$8nq_{p}(9) + 1 \equiv 8q_{p}(8) \rightarrow 16nq_{p}(3) + 1 \equiv 24q_{p}(2)$ (mod $p$).

The cases which relate $q_{p}(2)$ to $q_{p}(3)$ are merely some of the more interesting implications of the reformulation of Johnson's theorem suggested by your posting. Finally, if $p$ has the form $a^n -k$, where neither $a$ nor $k$ is divisible by $p$ nor greater than 89, a similar argument establishes the first case of Fermat's Last Theorem for $p$, since the failure thereof would require the vanishing of $q_{p}(a)$ for every $a \le 89$, as proved by Granville.