Generally speaking, when you get the error and the line causing the error is using a function call to pass its return value as parameter of another function, it means that the function getting the parameter is expecting a reference, while the function used to get the value to pass as parameter doesn't return a reference.

In your case, render() is defined as render(&$element), but node_show() is defined as node_show($node, $message = FALSE), not &node_show($node, $message = FALSE).
The same would be true for drupal_render(), since the function is defined as drupal_render(&$elements) and it expects a reference as its first argument.

As explained on Passing by Reference, when a function needs a reference as parameter, you can pass to the function:

• Variables
• The result of new
• References returned from functions

In the specific case, since you are using a function that is not returning a reference, you can only use a variable to store the result of node_show(), and pass that variable to render() or drupal_render().

Notice that, in 5.5.31 (and maybe also PHP 5.4), the Strict warning: Only variables should be passed by reference is not returned from the following code, which works as if test_array_print() obtained a reference.

function test_array_print(&$ref) {
  $ref[] = 10;
  print_r($ref);
}


function test_array() {
  return [3, 34];
}

test_array_print(test_array());

It prints the following.

Array
(
    [0] => 3
    [1] => 34
    [2] => 10
)

Still, see what Passing by Reference says:

No other expressions should be passed by reference, as the result is undefined. For example, the following examples of passing by reference are invalid:

<?php
  function foo(&$var)
  {
    $var++;
  }

  function bar() // Note the missing &
  {
    $a = 5;
    return $a;
  }

  foo(bar()); // Produces fatal error as of PHP 5.0.5, strict standards notice
              // as of PHP 5.1.1, and notice as of PHP 7.0.0

  foo($a = 5); // Expression, not variable
  foo(5); // Produces fatal error
?>