On the Pell-like $px^2-qy^2 = 1$ for prime $p,q$ Given any prime of form $p_n = u^2+nv^2$ for non-zero integers $u,v$. Consider,
\begin{aligned}
&p_2x^2-2y^2 = 1\\
&p_3x^2-3y^2 = 1\\
&p_7x^2-7y^2 = 1\\
&p_{11}x^2-11y^2 = 1\\
\end{aligned} 
Question 1: Are these four families of Pell-like eqns always solvable in non-zero integers $x,y$?
Ex.
Let $n=11$, and since $53=u^2+11v^2$, then $53x^2-11y^2=1$ is solvable. (Other $p_{11}$ will do.)
Question 2: Is there another $n$?
Let $n=13$, and since $157=u^2+13v^2$, but $157x^2-13y^2=1$ is NOT solvable in the integers.
Have these questions been answered in the literature already?  
 A: 1: Non-solvable equations for primes 2 and 3 
For prime 2,
$$97X^2-2Y^2=1$$
has no solutions and $97=5^2+2(6)^2$.  
For prime 3,
$$457X^2-3Y^2=1$$
there are no solutions and
$$457 = 5^2+3(12)^2$$
This should complete your list. You can check that a solution for either case means
$$X^2-194Y^2=97$$
or
$$X^2-1371Y^2=457$$
has a solution (just multiplying by a constant). This site can help you verify the non-solvability.  
2: Explicit results for prime 2 
The situation for
$$px^2-2y^2=1$$
is known. I have just posted a solution here.  
The results, quoting from there, is as follows:  
If $\beta\neq 0$ is an integer and $2\beta^2+1=\alpha^2p$ for some integer $\alpha$ and prime $p$, then the equation
$$px^2-2y^2=1$$
is solvable and one of the solutions is given by
\begin{align*}
x&=\alpha\\
y&=\beta
\end{align*}
This is easily checked:
$$px^2-2y^2=p\alpha^2-2\beta^2=2\beta^2+1-2\beta^2=1$$
(Note: The rest are obtained by composition with fundamental solutions.)  
Moreover, all solvable primes must occur in such a way. To solve this we can use the representation theorems of Binary Quadratic Forms, which is provided in the link.  
Note: In fact, this parametrized solution is a consequence of the representation theorems. It is not found by random guessing.  
Therefore you are looking for the intersection:
\begin{align*}
2\beta^2+1 &=\alpha^2 p\\
u^2+2v^2 &= p
\end{align*}
where $\alpha,\beta,u,v$ are integers. For your question you need $u^2+2v^2 = p$ implies $2\beta^2+1=\alpha^2p$ for some integers $\alpha,\beta$. This is not always true. For example, the equation I put on at the top
$$97X^2-2Y^2=1$$
has no solutions and $97=5^2+2(6)^2$.  
There is something similar for odd primes too, so that you can deduce conditions for the forms. However it is probably quite tricky to work out the details. (similar to what I did in my post)
