This answer is based on the excellent work of Will Jagy. This solves all cases of $k>3.$
Let $p<k$ be an odd prime such that $p\not\mid k.$
Solve $kd\equiv -1\pmod{p}.$ Let $n=(kd+1)/p.$ Note that since $p<k,$ $n>d.$
Then for any integer $t,$ we can take $z=a^{d}t^p$ so that $$\begin{align}az^k+1&=a^{kd+1}t^{kp}+1\\&=\left(a^nt^k\right)^p+1\\
&=(a^nt^k+1)\left(1+a^nt^k\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1}\right)
\end{align}$$
Where the last equation is because when $p$ is odd, $$
\begin{align}u^p+1&=(u+1)
\sum_{j=0}^{p-1} (-1)^ju^j
\\&=(u+1)\left(1+u\sum_{j=1}^{p-1}(-1)^ju^{j-1}\right)\end{align}$$
Now, since $n>d,$ we can set $$
\begin{align}x&=a^{n-d}t^{k-p}\\
y&=a^{n-d}t^{k-p}\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1}
\end{align}$$
For $k\geq 4$ we can always find such a $p$ by taking a prime factor of $n-1$ or $n-2$ if $n$ is even or odd, respectively.
So this solves all cases $k>3.$
You don't need $p$ prime, just that $1<p<k$ is odd and $\gcd(p,k)=1.$
k even
So when $k$ is even, we can take $p=k-1.$ Then $d=p-1$ and $n=p.$
Then for any integer $t,$ $$\begin{align}z&=a^{k-2}t^{k-1}\\x&=at\\y&=at\sum_{j=1}^{k-2}(-1)^j\left(a^{k-1}t^k\right)^{j-1}.\end{align}$$
k odd
Likewise, if $k=2m+1$ is odd, then you can take $p=2m-1,$ $d=m-1$ and $n=m.$ Then for any integer $t$:
$$\begin{align}z&=a^{m-1}t^{2m-1}\\
x&=at^2\\
y&=at^2\sum_{j=1}^{2m-2}(-1)^j\left(a^mt^{2m+1}\right)^{j-1}
\end{align}$$
is a solution.
In particular, for $k>3$ there are infinitely many solutions $(x,y,z)$ with $a\mid x$ and $x\mid y$ and $x\mid z.$
Claim
$ax + by = c$ has infinite number of positive solution if and only if $a$ and $b$ are of opposite signs and $\gcd(a,b)|c$
Proof
WLOG let $a>0$ and $b<0$. There exists a solution because $\gcd(a,b)|c$.
If $(x',y')$ is a solution then $(x'-b,y'+a)$ is a solution so we get a increasing ordered pair of $(x,y)$ that satisfies the equation. Hence there are infinite number of positive solutions for this case.
It is not possible if $a$ and $b$ are of same sign. Suppose there are infinitely many solution when $a$ and $b$ are of same sign then there exists a solution $(x',y')$ with $\mid x'\mid>c$ and $\mid y'\mid>c$ , but this implies $\mid ax'+by'\mid>c$ thus leading to a contradiction.
For example $(a,b,c)=(rt,-rt,2rt)$ where $r$ is a constant and $t$ is a integer variable will give you different values for $a,b,c$ such that the equation has infinitely many solutions.
Best Answer
The Diophantine equation to check on is
$$(x + y)(x + y + 2) = 10xy \tag{1}\label{eq1A}$$
Note that, by inspection, I got $(x, y) \in \{(0, 0), (0, -2), (-2, 0)\}$ being integer solutions, but not positive ones. Since there are integer solutions, doing things like factoring and checking for a perfect square discriminant will succeed (e.g., with $15x^2 - 10x + 1$, it's $1$ for $x = 0$ and $81 = 9^2$ for $x = -2$, with your Wolfram Alpha results show other, only negative, results)). One way to specifically show there are no positive integer solutions is to assume they exist but then use positive divisibility constraints (which assume all of the factors are non-negative), such as with my \eqref{eq4A}, \eqref{eq5A} and \eqref{eq6A} below, to show there are no valid results.
First, since $x + y$ and $x + y + 2$ have the same parity and $10xy$ is even, then both of the LHS factors of \eqref{eq1A} must be even. Thus, $x$ and $y$ have the same parity. They can't both be odd since the LHS has at least $2$ factors of $2$ but the RHS would have only $1$ factor of $2$. This means $x$ and $y$ must both be even. As such, there's a positive integer $d$ where
$$\gcd(x, y) = 2d \;\;\to\;\; x = 2dx_1, \; y = 2dy_1, \; \gcd(x_1, y_1) = 1 \tag{2}\label{eq2A}$$
Substituting this into \eqref{eq1A} and dividing both sides by $4d$ gives
$$(x_1 + y_1)(dx_1 + dy_1 + 1) = 10dx_{1}y_{1} \tag{3}\label{eq3A}$$
Since $d \mid (x_1 + y_1)(dx_1 + dy_1 + 1)$, but $\gcd(d, dx_1 + dy_1 + 1) = 1$, we have
$$d \mid x_1 + y_1 \tag{4}\label{eq4A}$$
Also, $x_1 + y_1 \mid 10d(x_{1}y_{1})$, but $\gcd(x_1, y_1) = 1$ means that $\gcd(x_1 + y_1, x_{1}y_{1}) = 1$, so
$$x_1 + y_1 \mid 10d \tag{5}\label{eq5A}$$
Note that \eqref{eq4A} and \eqref{eq5A}, along with \eqref{eq3A}, means there are positive integers $a$ and $b$ with
$$x_1 + y_1 = ad, \;\; d(x_1 + y_1) + 1 = bx_{1}y_{1}, \;\; ab = 10 \tag{6}\label{eq6A}$$
Substituting the LHS of \eqref{eq6A} into the middle part gives
$$\begin{equation}\begin{aligned} d(ad) + 1 & = b(ad - y_1)y_1 \\ ad^2 + 1 & = (ab)dy_1 - by_1^2 \\ by_1^2 - 10dy_1 + (ad^2 + 1) & = 0 \end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Treating this as a quadratic in $y_1$, the discriminant, i.e.,
$$100d^2 - 4b(ad^2 + 1) = 100d^2 - 4(10)d^2 - 4b = 60d^2 - 4b = 4(15d^2 - b) \tag{8}\label{eq8A}$$
must be a perfect square. Thus,
$$15d^2 - b \tag{9}\label{eq9A}$$
must also be a perfect square. From the RHS of \eqref{eq6A}, we get these cases to check for $b$:
Since $15d^2 - 1 \equiv 2\pmod{3}$, it's not a perfect square. In addition, if $d$ is odd, then $15d^2 - 1 \equiv 7(1) - 1 \equiv 6 \pmod{8}$, while if $d$ is even, then $15d^2 - 1 \equiv 3 \pmod{4}$, with neither being possible.
With $15d^2 - 2 \equiv 3 \pmod{5}$, it can't be a perfect square. Another method is to note that $d$ can't be even since it would have only $1$ factor of $2$, so $d$ is odd. However, then $15d^2 - 2 \equiv 7(1) - 2 \equiv 5 \pmod{8}$.
Checking $15d^2 - 5$, we get $d$ can't be even because then $15d^2 - 5 \equiv 3 \pmod{4}$. However, $d$ being odd means $15d^2 - 5 \equiv 7(1) - 5 \equiv 2 \pmod{8}$, which is also never true for perfect squares. Alternatively, $15d^2 - 5 = 5(3d^2 - 1)$, so $5 \mid 3d^2 - 1$. However, $d^2 \equiv 0, 1, 4 \pmod{5}$, with $5 \nmid 3d^2 - 1$ for each congruence.
Finally, $15d^2 - 10 \equiv 2 \pmod{3}$, so this can't be a perfect square either. Another way to show this is that $d$ can't be even because $15d^2 - 10$ would have only one factor of $2$, but $d$ being odd means that $15d^2 - 10 \equiv 7(1) - 2 \equiv 5 \pmod{8}$.
Since these all result in a non-perfect square discriminant, there are no positive integer solutions for $d$ and $y_1$ in \eqref{eq7A} and, thus, also for $x$ and $y$ in \eqref{eq1A}.