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.

Let $q$ be an odd prime such that ${q}^{t} + 1= 2{c}^{s} $, where $c, t$ are positive integers and $s= 1, 2$. We show that the Diophantine equation ${x}^{2} + {q}^{m} = {c}^{n} $ has only the positive integer solution $(x, m, n)= ({c}^{s} - 1, t, 2s)$ under some conditions. The proof is based on elementary methods and a result concerning the Diophantine equation $({x}^{n} - 1)/ (x- 1)= {y}^{2} $ due to Ljunggren. We also verify that when $2\leq c\leq 30$ with $c\not = 12, 24$, the Diophantine equation ${x}^{2} + \mathop{(2c- 1)}\nolimits ^{m} = {c}^{n} $ has only the positive integer solution $(x, m, n)= (c- 1, 1, 2). $