Elementary diophantine equations with unknown solutions Solvability of a general diophantine equation has been proved undecidable. 
As a famous example of knowledge, we know that $x^n+y^n=z^n$ has no solutions (in $\mathbb{N}$) for $n>2$. As a famous example of ignorance, we don't know whether or not $y=x+2$ has infinite prime numbers solutions.
Do you know some example of a simple Diophantine equation of which we ignore whether or not it has any solution? 
In particular is there some "famous" two variable Diophantine equation of unknown solvability? Maybe restricting possible solutions to prime, or to some subset of $\mathbb{N}$
*Basing on wikipedia $$2^{c-1}\equiv1\,(\mathrm{mod}\,c^2)$$
is not known to have composite solutions. This is the best I found so far.
 A: If you're asking for solutions where $x$ and $y$ is prime, there are none. $x^2$ diving $2^x$ immediately fails - note that if $x$ is prime, we have $2^x \equiv 2 \pmod x$ by Fermat's little theorem, and so $2^x - 1 \equiv 1 \neq 0$.
The equation you've given, $x^n + y^n = z^n$, has been solved for all naturals $n$ - there are trivial solutions for $n=1$, Pythagorean triads have been fully parametrised, and no solutions for $n > 2$ exist due to Fermat's Last Theorem, which was proven by Andrew Wiles in 1994.
An example of a simple statement which is still unsolved is whether infinitely many integers $n$ exist such that $n \mid 2^n + 3$. A more famous problem, although not Diophantine, is whether there exist infinitely many twin primes.
EDIT: the OP has since changed the equation from $2^x - 1 = yx^2$ to $2^x - 1 = yz^2$. In this case, it is unlikely that we can tell if there are infinitely many solutions - if we relax the prime condition on $z$ to allow $z=1$ the problem simply becomes finding Mersenne primes, for which there exists no simple algorithm to find as of current mathematical/computational techniques.
A: One of the simplest-to-state unsolved problems in mathematics is the Collatz conjecture:
Consider the algorithm:


*

*Choose a positive integer to start.

*If the integer is even, divide it by $2$.

*If the integer is odd, multiply by $3$ and add $1$.

*Repeat.


For every integer tried, following this algorithm eventually leads you to the cycle $4 \to 2 \to 1 \to 4\ ...$
The Collatz conjecture is that every starting integer will lead to this cycle. There might be some other repeating cycle out there, but none has ever been found.
