Let
$m$
,
$a$
,
$c$
be positive integers with
$a\equiv 3, 5~({\rm mod} \hspace{0.334em} 8)$
. We show that when
$1+ c= {a}^{2} $
, the exponential Diophantine equation
$\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(c{m}^{2} - 1)}\nolimits ^{y} = \mathop{(am)}\nolimits ^{z} $
has only the positive integer solution
$(x, y, z)= (1, 1, 2)$
under the condition
$m\equiv \pm 1~({\rm mod} \hspace{0.334em} a)$
, except for the case
$(m, a, c)= (1, 3, 8)$
, where there are only two solutions:
$(x, y, z)= (1, 1, 2), ~(5, 2, 4). $
In particular, when
$a= 3$
, the equation
$\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(8{m}^{2} - 1)}\nolimits ^{y} = \mathop{(3m)}\nolimits ^{z} $
has only the positive integer solution
$(x, y, z)= (1, 1, 2)$
, except if
$m= 1$
. The proof is based on elementary methods and Baker’s method.